Got a comment?
       
  Endgame 13 Jul 08: The surface with the larger area will depend on where the circle is drawn on the sphere. Let the radius of the circle be A cm, radius of the sphere be R cm. If A < (2^0.5)*R, area of spherical cap is greater than area of circle. If A = (2^0.5)*R, area of spherical cap is equal to area of circle. If A > (2^0.5)*R and A< 2*R, area of spherical cap is less than area of circle. If A = 2*R, area of spherical cap is equal to area of circle.

by:  Pramod Janardhanan     pramodj@lycos.com  on 17 July 2008 		   
  ANSWER TO THE PUZZLE: "How many rounds Gen.Goering's soldiers must be completing?" 1.I have received so far 4 answers(Kishore Rao/Dhruva Vaid/Shamik Banerjee/Vivek Jain),but none is right. 2.In the process of solving the problem,let's start by first asking:What precisely the Gen. had said? He had said: "your fattened BODIES have to move through five miles,nothing less..". 3.Next,let's critically analyse a "stride".When the soldier lifts his one foot,say left, and is on the point of putting it down (18 inches away),his right foot is still in the original "rest" position,though poised to be lifted.Thus when the FIRST stride is taken by the soldier,his "fattened body"(now onwards simply "body") has NOT moved through any distance AS YET. Now when the right foot commences moving forward,it momentarily comes ALONGSIDE the left foot(which is still on the ground),and then it rests on the ground,36 inches away from the original"rest"position. 4.It needs to be clearly understood that in the movement so far,the body has moved forth through only 18 inches;not 36 inches.If this is understood and accepted,the rest of the problem is simple. 5.Now, when the soldier is on the point of completing FIRST half round, he has just taken his 44th stride( which is with his right foot).In fact, completion of his 44th stride is marked by his RIGHT foot touching the south wall.But SIMULTANEOUSLY he also finds himself being obstructed by that wall!Which is when, as per the General's order, he rotates his heels through 180 degrees,makes an about turn,and is all poised to commence his SECOND half round.Well,he has taken 44 strides,but has his body moved through 44 strides? No !The body has moved through only 43 strides! 6.Now let's consider the SECOND half round.It is important to note here that, unlike in the first half,the body would be completing 42 strides when completing the SECOND half round ,which is when he is obstructed by the north wall.This is true of every subsequent half round, that is,the body would be moving through only 42 strides;not 44;not 43 either ! 7.Now, borrowing Mr Vaid's words,rest is simple arithmatic: a. First round...43 strides plus 42 strides = 85 x 1.5 ft =127.5 ft. b.For the body to move trough 5 miles=26,400 ft, the balance distance of 26,272.5 ft. remains to be covered. c.In EACH subsequent round,under the given conditions/constraints, the body moves through only 84 strides as seen above.Therefore,the distance moved(by"fattened bodies")is only 126 ft.and NOT 132 ft. d. At this rate,the no.of additional rounds required by "fattened bodies" to move through total of 5 miles, will be:26272.5 divided by 126 = 208 1/2 rounds, with a remainder of ONE stride. e. Lastly,add to this the FIRST round,which will give the answer: 209 rounds plus half a round plus one stride(18 inches). --Satyapal Sanghvi,9th July

by:  Satyapal Sanghvi     satyapal_sanghvi@yahoo.com  on 8 July 2008 		   
  This is my answer to Mr Satyapal Sanghvi's Puzzle of 12 June("How many rounds Gen.Goering's soldiers must be completing?") i)In the the puzzle,I found one point vague,so this is going to be a CONDITIONAL solution. ii)Since Gen. Goering utters:"left,right,left right",one may take it that the soldiers put-out their Left foot first,then right..so when the soldiers are JUST COMPLETING 44th STRIDE, their right foot will be on the point of touching the south wall.Which also means that soldiers now cannot continue their march further. But is it really so? iii)No!Because they can jolly well take their left foot forward,along side theirs RIGHT one, so that both the feet are just touching the south wall !But here comes in the sementics!The operative word given out by the General is :" Quick march". And in such "march",two feet cannot comeALONGSIDE,notwithstanding the fact that they all the while OUTPACE each other!So the left foot is not allowed to move further!The soldiers have to do about turn from this position. iV)Now,if these presumptions/restrictions are accepted by Mr Sanghvi as a problem-setter,the rest is a matter of simple arithmatic as follows: v)The distance between two walls being 66 ft, and a stride being of 18 inches,it takes 44 strides for soldiers to cover half a round or 88 strides to cover one round. Now 5 miles =5 x 5280 ft = 26,400 ft. So number of rounds= 26400 divided by 88 =300 rounds. vi)Is this the answer, Mr. sanghvi ? ---contributed by Dhruva Vaid (e-mail: dhruvavaid@gmail.com)

by:  Dhruva Vaid     dhruvavaid@gmail.com  on 5 July 2008 		   
  Another eror in the same (Genie) problem (A&B earning Rs30, not 36!)? Must be a record of sorts for MS! That of course doesn't make mine (equalising with the whole instead of half of the difference) any the less silly - the rap on the knuckle was well-deserved! Come to think of it, the problem may well be renamed the Jinx problem, looking at the pileup of errors. Cheers SD Pandit

by:  SD Pandit     sp242@rediffmail.com  on 27 June 2008 		   
  Orange problem:When the man sell for 20 ruppes. He sold the oange three ornge for one buck. The no is 30 no. The orange twoper buck is 20. Balnce orange remain whose cost two orange one buck. For 10 orange ge should get five bucks. As he sold 5 orange for 2 bucks. He will get four bucks which is less of one buck.He isnot stolen but selling in less price. submitted by adil_vnit@rediffmail.com

by:  Adil Mohmmed     adil_vnit@rediffmail.com  on 25 June 2008 		   
  Solution for Endgame 15 June 08: Area of the part between the 2 triangles is 6 sq cm. Let the length of the small triangle be A cm, length of the large triangle be B cm, radius of circle be R cm, area of small triangle be A1 sq cm, area of large triangle be A2 sq cm. For an equilateral triangle, area of small triangle is A1 = (3^0.5)*A*A/4 (eq1). The radius of the circle is related to the small side by R = A*(3^0.5)/3 (eq 2). The radius of the circle is related to the large side by R = B*(3^0.5)/6 (eq 3). Eq 2 and Eq 3 gives B = 2*A. Area of the large triangle is A2 = (3^0.5)*B*B/4 = ((3^0.5)*A*A/4)*4. Substituting from eq1, this gives A2 = 4*A1. Area of small triangle is given as 2 sq cm. Therefore, area of large triangle is A2 = 4*2 = 8 sq cm, which gives the difference in area as 6 sq cm.

by:  Pramod Janardhanan     pramodj@lycos.com  on 17 June 2008 		   
  I find something strange in archives. The columns for 25th May and 1st and 8th June are missing (all for 2008).This makes it difficult to backtrack answers.

by:  Avinash Saralkar     saralkar_avinash@yahoo.co.in  on 16 June 2008 		   
  These days, in which news paper, MS is getting published ?? As earlier it was in TimesOfIndia...

by:  Deepak Jain     deepakjain111@gmail.com  on 15 June 2008 		   
  testing.....

by:  Deepak     deepakjain111@gmail.com  on 15 June 2008 		   
  The solution to the e4 problem for 8 June 08 is: A is the Bottle Washer and earns 1890 sucks B is the Door Stopper and earns 2000 sucks C is the Door Opener and earns 1210 sucks D is the Doorknob Polisher and earns 1100 sucks E is the Welfare Officer and earns 1200 sucks The explanation is as follows: We first try to determine the worst paid employee. C is not the worst paid as C earns 10% more than the worst paid employee. A earns more than D, therefore A is not the worst paid. The Bottle Washer earns 0.945 times B’s earnings (from Clue 3), therefore B is not the worst paid. E earns 1.2 times Doorknob Polisher’s earnings less 100 sucks (from Clue 6), therefore E is not the worst paid, leaving D to be the worst paid. As the Bottle Washer earns 0.945 times B’s earnings and everyone earns in a multiple of 10 sucks, the only value which satisfies this criteria between 1000 and 2000 is either 1000 or 2000. 1000 is not valid as it would mean the Bottle Washer would earn below the minimum wage of 1000 sucks. Therefore B earns 2000 sucks and is also the Door Shutter (Clue 7). The Bottle Washer, therefore, earns 1890 sucks. From the paragraph above, neither C, D or E could earn 1890 sucks. Therefore, A earns 1890 and is the Bottle Washer. As C earns 10% more than D, and everyone earns in a multiple of 10 sucks, D must earn in a multiple of 100 sucks i.e. 1000, 1100, 1200 etc. As D is the worst paid, from Clue 4, E earns 100 sucks more than D. Therefore E must also earn in a multiple of 100 sucks i.e. 1100, 1200, 1300 etc. From Clue 5, C must be the Door Opener and earn an odd multiple of 10 sucks. For Clue 6 and 8 to be satisfied, E can earn either 1200, 1500 or 1800. E cannot earn 1500 as it does not satisfy Clue 5 and creates a sixth wage. E cannot earn 1800 as it creates a sixth wage. Therefore E earns 1200, D earns 1100 and C earns 1210 sucks. The Doorknob Polisher also earns 1100 sucks. Therefore D is the Doorknob Polisher and earns 1100 sucks. C is the Door Opener and earns 1210 sucks, leaving E as the Welfare Officer earning 1200 sucks.

by:  Pramod Janardhanan     pramodj@lycos.com  on 14 June 2008 		   
  May I present a puzzle to your readers through this "Message" space? The puzzle is presented in an anechdotal form,to add to it a little flavour !! PUZZLE Gen. Goering,the well-known Nazi Officer credited with being a legendary military disciplinarian(among other notorities),was fighting a last ditch battle.He with a handful of soldiers was holed up in a discarded ruin of what once might have been a concert hall,now left with just two walls amidst scattered heaps of debris. The men,who were starved for days together, luckily found good supplies of food and fed themselves well.Then,suddenly the General barked:" You have eaten like gluttons and fattened yourselves...But remember, you have to keep fit and trim..So now,let these fattened bodies of yours move through some miles and undergo some exercise.Its going to be Quick March for miles now..left right,left right..". A soldier, who couldn't suppress his ire, sarcastically asked: "But Sir,through how many miles these fattened bodies of ours can move when the distance between these two walls is just 66 feet ?" " You ..", the General muttered a choice expletive,and continued in the same streak of sarcasm:"Your fattened bodies have to move through full five miles...nothing less..", and completed with well-known German precision:"Start quick march from north wall..every stride of 18 inches..and when you reach south wall and can't move further,rotate your heels through 180 degrees,thus doing an about turn, and continue your march..reach north wall..and go on repeating likewise.. till 5 miles are over..". Now,would Mindsport readers like to work out how many rounds the soldiers must be completing when done ? --Satyapal Sanghvi,Mumbai 12 June 2008 (satyapal_sanghvi@yahoo.com)

by:  Satyapal Sanghvi     satyapal_sanghvi@yahoo.com  on 12 June 2008 		   
  18 where area would be 1296 times pi and volume would be 7779 times pi. For all other integers resulting in integers x pi results , either of the two would not be four digit integers. Ergo...

by:  Kishore Rao     kishoremrao@hotmail.com  on 9 June 2008 		   
  For my philosophical urgings--not the entire gamut but principally centred around "origins of humanistic values"-- I always looked around for some kind of interactions with like-thinking persons but was dismayed when I found them steeped in "conventioal" thinking,despite their professions otherwise ! And then comes along the indefatiguable name of Mukul Sharma!( Sychophancy by me? Sychophancy my foot! When a person, that is myself, has lived and believed in his own postulated theory on the subject, he has much to boast about and he just cannot be a sycophant !) For long I was thinking (ever since I switched to The Eco. Times) to contact you...and the trigger came when I read your piece"Where are the roots of morality?". I have postulated that all humanistic values/morals/sense of justice/rights/wrongs and the like are derivatives of how life CHOSE to evolve on this earth. That is to say :If "life",whatever be its definition, had evolved on a different PATH,all the above stuff would have evolved on altogether different lines ! I have my 3-page published article:"Impact of find of extra-terrestrial life--a philosophical speculation". Would Mukul Sharma be kind and courteous enough to let me mail to him my this article for his ERUDITE( no sychophancy ,indeed !)comment? -- Satyapal Sanghvi Mumbai, 7 June 2008

by:  Satyapal Sanghvi     satyapal_sanghvi@yahoo.com  on 7 June 2008 		   
  For my philosophical urgings--not the entire gamut but principally centred around "origins of humanistic values"-- I always looked around for some kind of interactions with like-thinking persons but was dismayed when I found them steeped in "conventioal" thinking,despite their professions otherwise ! And then comes along the indefatiguable name of Mukul Sharma!( Sychophancy by me? Sychophancy my foot! When a person, that is myself, has lived and believed in his own postulated theory on the subject, he has much to boast about and he just cannot be a sycophant !) For long I was thinking (ever since I switched to The Eco. Times) to contact you...and the trigger came when I read your piece"Where are the roots of morality?". I have postulated that all humanistic values/morals/sense of justice/rights/wrongs and the like are derivatives of how life CHOSE to evolve on this earth. That is to say :If "life",whatever be its definition, had evolved on a different PATH,all the above stuff would have evolved on altogether different lines ! I have my 3-page published article:"Impact of find of extra-terrestrial life--a philosophical speculation". Would Mukul Sharma be kind and courteous enough to let me mail to him my this article for his ERUDITE( no sychophancy ,indeed !)comment? -- Satyapal Sanghvi Mumbai, 7 June 2008

by:  Satyapal Sanghvi     satyapal_sanghvi  on 7 June 2008 		   
  Dear MS, History repeats.Remember,as originally stated, the sum and product problem had no solution.We fought several rounds blow to blow. For Edwina-Rustum problem there is a seemingly simple and clever answer.The form of the amount ,a perfect square ,can only be X00n with X odd and n even.n=6 is easily eliminated and it seems n=4. If Edwina keeps the change then the cheque is for 4.98 dollars. But still ,if one applies plain and solidly rigorous,undiluted logic,it turns out that there is no such integer.Use the fact that last p digits of root decide last p digits of the square.Only 498 and 998 sqare to form X004 .But unfortunately for the puzzle X is found even. Prefixing any digit to these candidates, such as 1498,2498,3498 etc. always throws up X as an even number. Hence there is no solution till kingdom come.

by:  Avinash Anant Saralkar     saralkar_avinash@yahoo.com.in  on 6 June 2008 		   
  Edwina should keep all the change of 4 cents and receive a cheque of 4.98 dollars. The total amount, a perfect,even square,has the form M0004 or M0006.The last two digits of any number,uniquely decide the last two digits of its square.No two digit number sqares to 06 combination in last places.Both these propositions can be easily computed or proven.One never has to know the total amount

by:  Avinash Anant Saralkar     saralkar_avinash@yahoo.co.in  on 4 June 2008 		   
  I have solved the e4 puzzle (the one with the clever mixed couple) but I really hope there is a better solution for this. Here it goes: Let total number of lamps be n and so each is sold at n cents. Therefore total money made is n^2 cents. Now, let number of $10 bills is x and number of 1 cent coins be y. Then, we have n^2 = 1000x + y which implies y can be from set [1,4,5,6,9] (perfect square) and also we know that x is odd. Rustam has 500 *(x + 1) cents and Edwina has 500*(x - 1) + y cents. I used C++ compiler to find smallest x and y which satisfy this. And they come out to be, x = 61, y = 9 and total money made = 61009 cents and number of lamps = 247. This means Rustam has 31000 cents and Edwina has 30009 cents. Since this money needs to be divided half and half, Rustam owes 495.5 cents to Edwina.

by:  Aseem Rastogi     r_aseem@yahoo.co.in  on 2 June 2008 		   
  hey MS why dont you come up with yet another site explaining the meaning of over complicated sentences which you use very often while describing your question,you could be a really good english teacher beleive me.do tell how should i develop same communication skills as yours

by:  Prahlad aggarwal     pa6339@gmail.com  on 1 June 2008 		   
  Hey Mukul, voice from the past .. have misplaced your contacts .. In Gurgaon till 28th of May , if you get this in time call my India MObile (0) 979 141 0035 .. hopefully we can meet for a coffee or a drink. regards, vish

by:  vish raju     vish_raju@yahoo.com  on 23 May 2008 		   
  I make it point to read your article in the E T and admire you for it.Admire you for your understanding of the subject matter and the way you express the same ,so dexterously. I suppose playing with words comes to you so effortlessly and spontaneously. It is gift. I did read an article penned by you for some NGO- about detention of jail inmates or about wrong treatment meted out to them etc,somehow it did not impress me and I dismissed you as one of the usual educated NGO cranks we have in our midst. I suppose we all have our idiosyncracies and your espousal of such causes should not cause me any consternation. Another great Sharma,is our Ruchir Sharma, any relation of yours ?.I make it a point to read his articles.He is too clever by half and if he thinks his oral dexterity can fool people to put their money where he suggests,he has another think coming.I have analysed his writings and have come to conclusion that he writes as directed, but in his own style. the thrust/direction/alledged newness and seminal ideas are all as directed by unseen Jew moneybag. Hats off to him for his verbal dexterity but I do not bracket you in that league. You are an angel compared to him and I like you. Your article on

by:  K K Luthria     luthriakk@gmail.com  on 21 May 2008 		   
  May 4, 2008 Endgames The Endgame really drove me crazy. After having spent a few hours on this, the closest I got to were the two words of CLASS and CRASS. The time could be 20: 18: 58 where the 0 becomes C in capitals, 1 becomes l in small letters, 8 becomes A in capitals and 5 and 8 become the letter S in captials. The search for the words was reduced significantly by the fact that the second and fourth digits could only be 0 to 5 (as an hour has only 60 minutes and a minute has only 60 seconds). This meant that the second and fourth letters could only be I, C, L, R and S. Also, the search was further restricted by the fact that only some letters could be represented by the digits.

by:  ravi kumar meduri     ravi_meduri@satyam.com  on 17 May 2008 		   
  pl provide ur contact details with ur personal cell no, pl. kindly ack

by:  San     m4@micronel.net  on 10 May 2008 		   
  Hi MS When we take the original sequence of 10,9,8,7,6,5,4,3,2,1,we end up with 7,3,1,5,9,2,4,6,8,10 in the pile with 10 at the bottom. 10 is in the right place we want so we leave it there. We want 9 where 8 has appeared, so we put 9 in the place 8 was in the original sequence and so on. Hence, we require 10,5,9,1,8,4,7,2,6,3,5,1 to end up with 1,2,3,4,5,6,7,8,9,10. As I have solved both e4 and endgame, you should print both... !

by:  Kishore Rao     kishoremrao@hotmail.com  on 30 April 2008 		   
  Hi, I would pick a sandglass as the time machine with the largest number of moving parts (grains of sand)and a sundial as the time machine with the smallest number of moving parts (none)

by:  Kishore Rao     kishoremrao@hotmail.com  on 30 April 2008 		   
  As minimax could be below or above the maximin, both possiblities are possible.

by:  Kishore Rao     kishoremrao@hotmail.com  on 23 April 2008 		   
  Hi MS, The e4 is a brilliant observation from a six yo. When only the upper half is seen, no changes are visible when moving from 2 to 3, 5 to 6 and 8 to 9 (as the changes happen in the eclipsed lower half), whereas when the whole display is visible changes continously. Presuming the number increments every seconds, when seeing the upper half only, 7 changes happen in 10 seconds as against 10 changes in the same period when viewing the whole. Shows, that the clock ticketh faster for those who see the whole picture....

by:  Kishore Rao     kishoremrao@hotmail.com  on 23 April 2008 		   
  Hi! This is solution to end game 20 April 2008. Let there be a matrix of people standing in 20 columns and 10 rows. Let person

by:  Yogesh Chandola     yogesh.chandola@gmail.com  on 22 April 2008 		   
  Answer for Puzzle-"U GUEST RIGHT" are: Andrew Soames Jon Newton & Lisa Newton Bea Vogel & Ralph Vogel Kathy Duval Veronica McCathy.. Apprise with your comments.

by:  chintan     chintan_maverick@yahoo.co.in  on 17 April 2008 		   
  What an engrossing e4 this week. I spent most of time in my office solving this one and I hope it's correct :) (also, that my boss isn't reading). Here is the solution to McHannot's dilemma: Mr. Ralph Vogel arrives with wife Mrs. Bea Vogel on 20th August (Monday). On 21st, Ms. Lisa Duvel arrives. On 23rd, Ms. Lisa Duvel leaves after spending two nights. On 24th, Mr. & Mrs. Vogel leave after spending 4 nights and Mr. Andrew Newton with his wife Mrs. Kathy Newton arrives to the crazy place. On 25th (Saturday), Mr. Jon Soames, the happy single man, arrives. 26th goes without any arrivals and departures. On 27th, Mr. Jon Soames departs and Miss Veronica McCarthy arrives. On 28th, Mr. Andrew Newton departs with his wife Mrs. Kathy Newton after spending 4 nights. After a quite 29th and 30th, Miss Veronica McCarthy leaves after she had spent 4 nights at the place and McHannots heave a sigh of relief. Phew !!!!

by:  Aseem Rastogi     r_aseem@yahoo.co.in  on 14 April 2008 		   
  My previous mail seems to have lost formatting, so here goes again: I could come out with this: A B C - 13 7 7 13 - 19 1 - 12 1 7 12 8 - 5 8 7 5 13 2 18 - 2 18 2 - 11 2 7 11 9 - 4 9 7 4 13 3 17 - 3 17 3 - 10 3 7 10 10 - Maybe there is a faster way, but it may require more containers ! I am sure all the present jugs could not have a shorter number of iterations. I could come out with this: A B C - 13 7 7 13 - 19 1 - 12 1 7 12 8 - 5 8 7 5 13 2 18 - 2 18 2 - 11 2 7 11 9 - 4 9 7 4 13 3 17 - 3 17 3 - 10 3 7 10 10 - Maybe there is a faster way, but it may require more containers ! I am sure all the present jugs could not have a shorter number of iterations.

by:  Kishore Rao     kishoremrao@hotmail.com  on 10 April 2008 		   
  I could come out with this: A B C - 13 7 7 13 - 19 1 - 12 1 7 12 8 - 5 8 7 5 13 2 18 - 2 18 2 - 11 2 7 11 9 - 4 9 7 4 13 3 17 - 3 17 3 - 10 3 7 10 10 - Maybe there is a faster way, but it may require more containers ! I am sure all the present jugs could not have a shorter number of iterations.

by:  Kishore Rao     kishoremrao@hotmail.com  on 10 April 2008 		   
  Dear MS, I beg to state a sincere request from a few poor chaps including me. Could you please update this website on Sundays so that we can get the required time to attempt the questions without caring much for our professional demands? Thanking you very very much in anticipation. Yours truly, Shamik Banerjee

by:  Shamik Banerjee     banerjeeshamik123@yahoo.com  on 6 April 2008 		   
  Enjoyed your write-up on Frill technology in todays TIO. a brilliant write-up indeed ! Arvind -a fan of yours since time immemorial.

by:  Dr.Arvind Mishra      drarvind3@gmail.com  on 1 April 2008 		   
  Hi MS, I am in a perpetual bind with your e4, today, I feel like a gramophone with a stuck needle, singing "I am nineteen, going on nineteen, going on nineteen ...." As you estimated my age at 900. Your estimate was way off.

by:  Kishore Rao     kishoremrao@hotmail.com  on 1 April 2008 		   
  The words given, when pronounced, give out the sound of the alphabet (in their order); though the alphabet are not actually present in these words. We can write the words as follows : bouq-A, B-eijing, C-ling, D-aoism, qwe-E, F-armacy, G-initor, nava-H-o, I, J-esture, K-orus, dub-L-u shaped, gra-M-pa, co-N-troller, tabl-O, hiccu-P, Q, co-R-nel, S-enturion, pass-T, U, thero-V, W-on, wre-X, Y-ise, Z-ylophone. Saifuddin Khomosi, Dubai.

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 24 March 2008 		   
  How do I search you site, if I know part of the question?

by:  Thirumoorthy     send2moorthy@indiatimes.com  on 18 March 2008 		   
  Hi, This is test messages posted from http://www.mindsport.org/writetoms.html Thanks

by:  Deepak     deepak@roseindia.net  on 17 March 2008 		   
  phew! testing again

by:  babu     bv@kaledezign.com  on 17 March 2008 		   
  Suman Comment posting

by:  suman     suman@roseindi.net  on 15 March 2008 		   
  Testing from write to MS

by:  babu     bv@kaledezign.com  on 15 March 2008 		   
  testing from comments page

by:  babu     bv@kaledezign.com  on 15 March 2008 		   
  testing

by:  babu     bv@kaledezign.com  on 15 March 2008 		   
  This is also a test

by:   mukul sharma     mukul.mindsport@gmail.com  on 14 March 2008 		   
  This is a test message

by:  Binita Mohanty     binitamohanty@yahoo.com  on 14 March 2008 		   
  Hi MS The answers are ADDER CHAIN FILLS MILLS COBRA BANJO CHEER SNEER SHEER PECAN I used MS Excel and created rows linked to a starting row. In the first row I entered the ASCII code of the clue. In subsequent rows I used the function CHAR along with clue -1 ASCII code and so on and just read down the rows. Actually took less than 10 minutes, since quite a few were within the first few rows and quite a few had common letters. Thanks for printing my answer. Kishore

by:  KISHORE RAO     kishoremrao@hotmail.com  on 12 March 2008 		   
  Could decipher the words in a jiffy. Here's the solution : BEEFS = ADDER ; INGOT = CHAIN ; LORRY = FILLS ; SORRY = MILLS ; FREUD = COBRA ; FERNS = BANJO ; JOLLY = CHEER ; TOFFS = SNEER ; TIFFS = SHEER ; TIGER = PECAN.

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 12 March 2008 		   
  Soln for the Endgame on 11th Nov 2007: The minimum number that turns out to satisfy professor infinities quest assuming all numbers involved are integers is 5184 as the total amt to be inherited (do we assume its in millions?? then are the daughters available coz theres two vella guyz here who are very interested). The number basically is 2^6 * 3^4 = 5184. accordingly each daughter inherits 1330 million $.... now if the answer is any multiple of 5184 then we are even more interested...n the vella part well we are here solvin Mindsport in office arent we...need i say more...

by:  Nirmal Kumar Iyer and Mehul Dixit     nirmal.iyer@gmail.com, mehul.m.dixit@gmail.com  on 10 March 2008 		   
  Solution for SLAPPIN' LEATHER posted on 3rd Feb 2008: Bat Masterson's average hit rate would be 81/108 which turns out to be 75%. Now if Muruggan oops Morgan Earp misses 25 cons shots then his hit rate turns out to be 75/100 which of course if 75% again... so Mr Bat aint right...but in hindsight for Mr Muruggan to quit at 25 missed shots was very fortunate indeed...coz for muruggan to gun 6/8 shots to equal Batman was quite improbable considerin the inebriated state that he was in...

by:  Mehul Dixit     mehul.m.dixit@gmail.com  on 10 March 2008 		   
  The problem of Rohan, Sharjah is indeed simple. Let us consider a square ABCD with side X. On one side, say AD, make a vertical cut at X/2 (say at E going up to F). Pick up one of the pieces (EFCD) and make a diagonal cut (CE) to get two identical right-angled triangles (CEF and CED) with X and X/2 as non-hypotenuse sides. Now it is now a just a question of rearranging. To get rectangle, place both the triangles CEF and CED alongside the rectangle ABEF. To get parallelogram, place triangle CEF on the side of ABEF with CF touching BF. Similarly for the triangle CED on the other side with DE touching AE. Just flip one of the triangles and you get trapezium. For the right-angled triangle, keep the triangle CEF on the side ABEF with CF touching BF. Keep the other triangle CEF on top of ABEF with CD touching EF. Mohan

by:  NMOHAN     nmohan@omantel.net.om  on 7 March 2008 		   
  February 17, 2008 Hmmm. I appreciated the retro comments made by those oldtime Mindsportians (are they still actively having fun?) only when I neared the completion of the crossword. So we have two solutions based on whether the two daughters of Farmer Dunk are twins are not. In solution 1, the Area of Dog’s Mead is 38720 square yards (220 by 176 yards); farmer dunk is 72 years old but can still walk at 27 mph; his mother in law 86 years old; Edward 45 years old; Martha 32 years old and Mary who was born in the year 1913, 22 years old. The pigley farm was first occupied in the year 1610 and is in possession of the Dunk family for 326 years. The Dog’s Mead portion of the farm costs 355 shillings an acre. In solution 2, the two daughters, being twins, are of the same age 22 years. All the other values remain the same as solution 1, except for a) the age of Mrs. Grooby, who would be 85 years old b) the year of first occupation, which would be 1510 c) Birth year of Mary, which would be 1914 d) Period of possession of the pigley farm, which would be 426 years

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 7 March 2008 		   
  The solution given for the Divine Proportions Dept matches only 4 elements of the two triangles, where as the solution given by me matches all 5 elements as the original problem stated.

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 6 March 2008 		   
  March 1, 2008 ENDGAME Make the cuts along the two diagonals of the square to get four equal right angled triangles. The triangles can then be arranged to get the rectangle, the parallelogram, the trapezium and the right angled triangle.

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 6 March 2008 		   
  March 1, 2008 ENDGAME Make the cuts along the two diagonals of the square to get four equal right angled triangles. You can triangles can then be arranged to get the rectangle, the parallelogram, the trapezium and the right angled triangle.

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 6 March 2008 		   
  By cutting along one diagonal and also along the line joining the midpoints of two opposing sides, we can make a rectangle, two difference parallograms, a trapezium and a right angled triangled as given in attached file. We can also make an irregular hexagon too.

by:  Kishore M     kishoremrao@hotmail.com  on 5 March 2008 		   
  hi First cut should be vertical or horizontal dividing the square into two equal rectangles. The second cut should divide diagonally one of these rectangles into two equal right angles |\ |\| ||\ /||/ | | /||\ Thus in this way with these two triangles and one rectangle wecan fornm all the four shapes Rajeev K Mital

by:  Rajeev mital     rajeev_mital@hotmail.com  on 4 March 2008 		   
  aa

by:  aaa     aa@aa.com  on 2 March 2008 		   
  The snow started falling at 11:20 am. Assume the snow is falling at h miles/hr and the snow has been falling for x hrs prior to 12 pm. Therefore at the end of the first hour, the relation between rate of travel of the plough and height of snow is 1=k/(xh+h), where k is constant of proportionality. Likewise at the end of the second hour, the relationship is given by 1.5/2 = k/(xh+2h). Solving the two eqns gives x = 2/3 hrs or 40min before 12pm i.e. 11:20 am Regards Pramod

by:  Pramod Janardhanan     pramodj@lycos.com  on 29 February 2008 		   
  February 10, 2008 ENDGAME If it were to be from the ground, then one would have to throw the ball from another 20 feet from the wall at a velocity of root of 100g and at an angle of tan-12 or about 63.50. However, since the ball is being thrown from a height of 5 feet, the velocity is slightly lesser at 54.2 feet per second and the distance from the wall can be slightly lower at 17.42 feet from the wall. One can solve this problem by first assuming that the stone is thrown from the level of the ground and then by applying the correction for the height of five feet. This can be done by extrapolating the position and the velocity of the stone when it reaches the height of five feet. For solving the stone thrown from the ground, one has to combine the two equations of VSinǾ*d/VCosǾ - ½ g * d2 / (V2*Cos2Ǿ) = 30 (the vertical clearing) 2 VCosǾ VSinǾ / g = 40 + 2d (the horizontal clearing) Where V is the velocity with it has to be thrown, Ǿ the angle with the ground, g the deceleration due to gravity and d the distance from the wall. The combined equation expressed in terms of just V and Ǿ can be differentiated and dv/dǾ equated to zero. e4 Thanks for the problem that made me search for the integral of 1/x. Let the speed of the snow plough be V cubic miles per hour of snow being cleared. Let snow fall uniformly at X miles per hour. Let the breadth of the road be B miles. At any point of time, the horizontal speed of the snow plough = V / (B*Height of Snow) = V / (B*X*t) where t is the duration for which snow has fallen. If we assume that the snow started falling T hours before noon, the initial horizontal velocity of the snow plough is V / (B*X*T) or say K / T where K is a constant equal to V / (B*X). Consider a small interval of time dt after the plough has started. The horizontal velocity at this time t = K / (T + t) The horizontal distance travelled in the first hour would therefore be 0∫1 K / (T+t) dt = 1 mile gives the value of K as 1 / log [(T+1)/T] 1∫2 K / (T+t) dt = 0.5 mile gives the value of K as 0.5 / log [(T+2)/(T+1)] Equating the two K values gives the value of T as approximately 0.75 hours or about 45 minutes. Thus the snow started falling approximately at about 11.15 a.m.

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 29 February 2008 		   
  hello sir, i am a big fan of mr mukul sharma's articles i searched on internet but didn't found them could you please provide me a copy of them(all the articles). thanking you in anticipation. nakul sharma

by:  nakul sharma     indian.nakul@gmail.com  on 28 February 2008 		   
  Dear MS, I am currently attending a programme on

by:  Col Sravan Kumar     sravan.marur@gmail.com  on 25 February 2008 		   
  RE-MINDSPORT : 17/2/08 Area = 38720 sq.yds.; Martha's age = 32 yrs.; Year of farm occupation = 1610; Farmer Dunk's age = 72 yrs.; Mary's birth = 1913; Mrs.Grooby's age = 86 yrs.; Mary's age = 22 yrs.; Value of Dog's Mead = 142 pounds; Edward's age = 45 yrs. The farm has been in the family's possession for 325 yrs. Whew ! By the way where is Dog's Mead located ? SAIFUDDIN KHOMOSI, DUBAI.

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 20 February 2008 		   
  RE-MINDSPORT (17/2/08) Hooray ! I did it. Sure was a difficult one ; but quite engrossing. Came across such a googly (or Pigley rather) after a long period.The solution is as follows : 3 8 7 2 0 X 1 ; 5 X 3 2 X 4 4 ; 5 X 9 X 3 5 2 ; X 1 6 1 0 X X ; 7 2 X 1 9 1 3 ; 9 X X X 7 9 2 ; 2 7 X 1 6 X 5. Area=38720 sq.yds.; Martha

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 20 February 2008 		   
  CUE BRUTE every cube has eight pieces with three sides coloured. the smallest cube which has one uncolored cube; has three units. if the number of uncolored cubes is 64, they form a 8 unit cube, which is inside a colored cube, so the colored cube must have 10 units each edge. q.e.d.

by:  vivek gharpure     callisto@sify.com  on 20 February 2008 		   
  Dear MS, Thanks for giving me the pleasure of solving this puzzle (2/17/08) all over again - I don

by:  Chaitra D     chaitra.dk@gmail.com  on 19 February 2008 		   
  Throwing Stones - Endgame (10 Feb 2008) The most efficient projectile for maximum horizontal distance travelled with least energy is when projected at an angle of 45 degrees. Thus the stone is thrown from a distance of 37.016 feet at a velocity of 60.407 ft/sec to just clear the top of the rather thick wall.

by:  Ben Cooper (Muscat, Oman)     bencooper16@hotmail.com  on 17 February 2008 		   
  testing MS!

by:  Babu Vennikal     bv@kaledezign.com  on 16 February 2008 		   
  How does a syndicated columnist (or pagist) deal with a howler? Pray that no one noticed it. If drawn attention to, admit in a scale of “gracefully to sheepishly”. Worst-case handling is to draw further attention by giving a lengthy unrelated example. I missed your original howler completely but was drawn to it fully by your impish explanation. The prefix of light in front of a unit of time like year is to easily measure and share astronomical distances. Period. You just picked up the phrase ‘light second’ (instead of milli, micro or nano) in a hurry without realizing that it is nothing but a simple second if one is to describe time. So why not just accept it and move ahead? Mohan

by:  Mohan     deepak@roseindia.net  on 16 February 2008 		   
  How does a syndicated columnist (or pagist) deal with a howler? Pray that no one noticed it. If drawn attention to, admit in a scale of “gracefully to sheepishly”. Worst-case handling is to draw further attention by giving a lengthy unrelated example. I missed your original howler completely but was drawn to it fully by your impish explanation. The prefix of light in front of a unit of time like year is to easily measure and share astronomical distances. Period. You just picked up the phrase ‘light second’ (instead of milli, micro or nano) in a hurry without realizing that it is nothing but a simple second if one is to describe time. So why not just accept it and move ahead? Mohan

by:  NMOHAN     nmohan@omantel.net.om  on 15 February 2008 		   
  February 3, 2008 e4 The e4 of February 3, 2008 has two solutions based on the rules decided by Deputy Sheriff Doc Holliday. Generally in sports if you do take your chances they are considered to be forfeited and you score zero on those chances. If this is the rule to be applied, then Bat has a better record (81 on 108) against Morgan (75 on 108). If the rounds that Morgan has given up are not to be counted, then both have the same record of 75% accuracy (81 on 108 and 75 on 100). In this case, given the form that Morgan was in after filling up, Bat should have insisted that Morgan complete all his rounds. ENDGAMES Dear MS, There is an easier way out than the easy way out given by you and that is to solve the ENDGAME itself. It was so mindboggingly simple that I was shocked that this has come directly from you. Anyway, the number is 3, the consecutive powers are 3^2 (9) and 3^3 (27) and the permutation is 729 which is a higher power of 3^6.

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 15 February 2008 		   
  e4 The e4 of February 3, 2008 has two solutions based on the rules decided by Deputy Sheriff Doc Holliday. Generally in sports if you do take your chances they are considered to be forfeited and you score zero on those chances. If this is the rule to be applied, then Bat has a better record (81 on 108) against Morgan (75 on 108). If the rounds that Morgan has given up are not to be counted, then both have the same record of 75% accuracy (81 on 108 and 75 on 100). In this case, given the form that Morgan was in after filling up, Bat should have insisted that Morgan complete all his rounds

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 15 February 2008 		   
  ENDGAMES The sheet of paper with numbers problem on January 13, 2007, needs lateral thinking which is tough for people like me. Any way, giving it a try they could be answers to ten questions posed in an exam paper in that order. Don’t ask what those questions are. They are those which have answers of 100, 32, 30 etc. in that order. Or they must be the results of a holistic assessment of the student including his height, weight, absenteeism etc.

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 15 February 2008 		   
  The locks problem (December 30, 2007) though seemed tough initially was in the end simple enough. The three conditions result in the following deductions: 1) Each pair of slaves must have atleast one lock that they cannot open (5C2 = 10 such pairs). To have minimum locks the locks should have fewer keys. Foe example, if locks have only two keys given to slaves, then there are 3 pairs that cannot open that lock 2) Any threesome of slaves must be able to open all the locks implies that every lock must have atleast three keys given to three different slaves. Otherwise, there will be a combination of three slaves that cannot open that lock. Combining this with the above, the number of locks are minimized if each lock has three keys given to a unique threesome (there are ten such threesomes 5C3 = 10) 3) The King and any slave must open all the locks means that the King must have the key of all locks except for one and that one lock can be opened by any slave Thus we will need a minimum of 11 locks to create such an arrangement. For example an arrangement is given below: Lock 1 – 5 keys to all slaves Lock 2 – Keys to King, 1st, 2nd and 3rd slave Lock 3 – Keys to King, 1st, 2nd and 4th slave Lock 4 – Keys to King, 1st, 2nd and 5th slave Lock 5 – Keys to King, 1st, 3rd and 4th slave Lock 6 – Keys to King, 1st, 3rd and 5th slave Lock 7 – Keys to King, 1st, 4th and 5th slave Lock 8 – Keys to King, 2nd, 3rd and 4th slave Lock 9 – Keys to King, 2nd, 3rd and 5th slave Lock 10 – Keys to King, 2nd, 5th and 4th slave Lock 11 – Keys to King, 3rd, 4th and 5th slave

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 15 February 2008 		   
  This is a test message

by:  Binita Mohanty     binitamohanty@yahoo.com  on 14 February 2008 		   
  Birdie in a hole. Something kept bugging me that the solution had some vague golfing connection. Then it struck ! a rainy day, a flooded green and out pops the bird (birds float, or at least sea-gulls do).

by:  Ben Cooper (Muscat, Oman)     bencooper16@hotmail.com  on 12 February 2008 		   
  Consecutive Powers. Following the big build-up (sky-high neuro-medical bills etc) I thought of 10 digit numerical anagrams and dragging-up good old c++ .... too much effort, so I doodled on the back of a fag packet for a few seconds. To my amazement I found 2 solutions for digit 3 - powers 2 and 3, and 3 and 4 give anagrams of 729 and 2187 which are 6th and 7th powers of 3 respectively. However, somewhere out there, there must be an elusive 10 digit anagram........

by:  Ben Cooper (Muscat, Oman)     bencooper16@hotmail.com  on 12 February 2008 		   
  what a show!

by:  Ben Cooper     bencooper16@hotmail.com  on 12 February 2008 		   
  The answer to 7 x7 crossword (e4 Feb 10, 08) is as follows : 4 8 6 x 4 9 6; 8 x 4 6 x 1 8; 3 2 1 9 2 5 7; x 1 8 1 x 2 x; 6 9 4 x 3 3 3; 9 x 4 4 1 x 6; 7 7 6 x 2 2 5; . I could get it in ten minutes time. The first step was to fit in seven digit nos. across & down (with only one pairing possible becos only no.1 is common in both). The second step was to fit the five digit no. Then there was smooth sailing all the way (fitting in bits & pieces).

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 11 February 2008 		   
  The answer to 7 x7 crossword (e4 Feb 10, 08) is as follows : 4 8 6 x 4 9 6; 8 x 4 6 x 1 8; 3 2 1 9 2 5 7; x 1 8 1 x 2 x; 9 x 4 4 1 x 6; 7 7 6 x 2 2 5; . I could get it in ten minutes time. The first step was to fit in seven digit nos. across & down (with only one pairing possible becos only no.1 is common in both). The second step was to fit the five digit no. Then there was smooth sailing all the way (fitting in bits & pieces).

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 11 February 2008 		   
  The answer to 7 x7 crossword (e4 Feb 10, 08) is as follows : 4 8 6 x 4 9 6 8 x 4 6 x 1 8 3 2 1 9 2 5 7 x 1 8 1 x 2 x 6 9 4 x 3 3 3 9 x 4 4 1 x 6 7 7 6 x 2 2 5 I could get it in ten minutes time. The first step was to fit in seven digit nos. across & down (with only one pairing possible becos only no.1 is common in both). The second step was to fit the five digit no. Then there was smooth sailing all the way (fitting in bits & pieces).

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 11 February 2008 		   
  Hi Old timers who have used a typewriter might remember that the ink ribbon usually had double the width that was actually required. Some ribbons had the top half of width of the ribbon with one colour and the other half with another colour. If you wanted a single colour, that too was available and you could use one half of the width till it started to run out of ink and switch to the other half and use it up. Since this had the problem that when you were using the first half, the second half was still drying due to exposure to air, some people even switched sides daily, since a convenient switch was available for this, so that both halves got used to fairly similar levels. With the change to dot matrix printers, the ribbons became continuous loops and had a width to the extent required, so that the ribbon endless looped around. But suppose you had a printer ribbon cartridge which could accomodate a ribbon of double width and you had a spool of such a ribbon width from the typewriter days, what would be best method of fitting it into the cartridge (with addition of suitable guides, if required) so that both the halves could be used evenly, keeping in mind that no switch is available to switch between the two halves? Further challenge for Hindi song afficionados: Give me the answer through the lines of a hindi song lyric Kishore

by:  Kishore M     kishoremrao@hotmail.com  on 10 February 2008 		   
  Hi MS, one solution is 3^2=9,3^3=27,3^6=729. Also,3^3=27,3^4=81,3^7=2187. I have tried only the low powers of digits, but in the higher powers? well in mathematics sky is the limit and we have not plumbed the depths (?) of the sky. So who knows what exists out there.. Kishore

by:  Kishore M     kishoremrao@hotmail.com  on 9 February 2008 		   
  With reference to the Endgame about "what two consecutive powers of a digit are such that....", the digit is 3 and the powers are 3 and 4 3^3 = 27 and 3^4 = 81 on permutation of combined digits you get 2187 which is 3^7. in spite of all the dire warnings, i managed to crack it in 8.271 secs (another permutation... would that be coincidence??)

by:  Shreni Raj     skywise3000@hotmail.com  on 9 February 2008 		   
  Solution for the Endgame: There are at least two answers: (1) Second and third powers of 3 (i.e., 9 and 27 respectively) when combined, produce the sixth power of 3 (= 729). (2) Similarly third and fourth powers of 3 (27 and 81 resp.) produce the seventh power of 3 (2187). - Sheikh Sintha Mathar, sheikhsm@eim.ae

by:  Sheikh Sintha Mathar     sheikh@alrostamanigroup.ae  on 9 February 2008 		   
  Endgame Feb 3,'08 : The digit 3 raised to the powers of 3 & 4 gives 27 & 81 respectively. On permutating 27 & 81, we get 2187 (which is 3^7).

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 6 February 2008 		   
  Endgame (Feb 3,

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 6 February 2008 		   
  The answer for this endgame is 3^2 =9,3^3=27 permuting these two we have 729 which is 3^6.

by:  PVS murthy     frommurthy2u@gmail.com  on 5 February 2008 		   
  Endgame Jan 27 : How about pouring sand along the sides slowly and filling the cavity gradually. As the level of sand rises, the bird will help itself up. Reminds one of the thirsty crow & the water pot ,eh?

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 4 February 2008 		   
  The newer series of BOEING airplanes have the ends of their wings turned upwards - for a foot or two . Does this serve any specific purpose / help in aerodynamics ?

by:  Vivek Jain     vivekdr@gmail.com  on 3 February 2008 		   
  Which newspaper is mindsport published? What other articles you write regularly & in which periodicals / newspapers? Please reply on my mail ID. I would be gratefl.

by:  MILIND KSHIRSAGAR     milindkshirsagar_2008@rediffmail.com  on 30 January 2008 		   
  ENDGAME 27th JAN; A.Pour water ( slowly ) into the hole. B. Drop in a thread with worms at regular intervals on it - measure its depth . Next lower a "L" shaped rod alongside the wall such that the short arm of the rod can be expanded as it reaches the bottom ( a stent like thing ) and then lift up the bird.

by:  Vivek Jain     vivekdr@gmail.com  on 30 January 2008 		   
  ENDGAME 27th JAN; A.Pour water ( slowly ) into the hole. B. Drop in a thread with worms at regular intervals on it - measure its depth . Next lower a "L" shaped rod alongside the wall such that the short arm of the rod can be expanded as it reaches the bottom ( a stent like thing ) and then lift up the bird.

by:  Vivek Jain     vivekdr@gmail.com  on 30 January 2008 		   
  btw, do we post our answers here? the last time i'd posted was in sunday times!!!

by:  Devansh Patel     devansh.dash@gmail.com  on 29 January 2008 		   
  answer to SELLING DOWNSTREAM as usual, the question is not tough. nothing ever is. its jus the way it has been presented... and in turn the way you look at it, makes it appear complicated. the statement is quite true. the boat drifting along can be steered only if its faster- reason being, steering is effective only if there is relative motion between the boat and the water flow. also since the boat can not be slower than the stream pushing it and earlier, as mentioned, had been powered, will have some momentum of its own, making it go faster than the stream, also accounting for the relative motion explained earlier. i think language is more complicated than such trivial questions themselves... what say you? ;)

by:  Devansh Patel     devansh.dash@gmail.com  on 29 January 2008 		   
  Endgame Jan.27,08 : Well how about putting a grain (or something) tied to a string inside the hole, so that the bird can catch it in its beak & be pulled out safely ?

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 29 January 2008 		   
  Any cube has 8 corners (each having 3 painted faces), 12 edges (each with n-2 cubes having 2 painted faces) and 6 faces (each with (n-2)squared cubes having 1 painted face). The interior (n-2)cubed cubes have no face painted. "n" here stands for the dimensions of the original cube (in inches). The answer to the problem is n = 6 inches. This cube, when cut, will form 96 cubes with one painted face which is twice of 48 cubes haing two painted faces. Also, 64 cubes with no face painted would be formed which is eight times the 8 corner cubes having 3 painted faces.

by:  Ajay Chauhan     chauhan.ajay@gmail.com  on 27 January 2008 		   
  Dear Mukul, Answer to question on wall clocks submitted by Kartik. I am not sure whether this is the concept used. But this can be used to appropriately compensate for the additional potential energy on account of gravitation. In the hand wound clocks, potential energy is stored in the spring, which on unwinding provides the kinetic energy of movement. The potential energy due to gravitational force should not interfere with it. As the hand moves down, the gravitational potential energy should be made to wind up a spring through an appropriate mechanism so that it does not provide any additional kinetic energy to the hand movement. The same stored potential energy should be released when the hand moves up to nullify the gravitational force acting against the upward movement. This is okay as far as wall clocks are concerned. There were smaller wrist worn watches. I am not sure what type of mechanism, if any is needed for them. May be the effect is miniscule for such tiny mass. Mohan

by:  N.MOHAN, TOYOTA, OMAN     nmohan@omantel.net.om  on 25 January 2008 		   
  Mr Mukul, Hope you are well. I was just awaiting your reply on my question. Thanks

by:  Shadab Malik     malik112@gmail.com  on 24 January 2008 		   
  Dear Mukul, When i cut and pasted for Cue Brute (Mindsport20th jan), the last para got left out, though the answer I pasted was complete in itself. If you feel appropriate, you can append this last para too! "We can look at a general solution as well. For a cube of dimension X units (obviously for X more than one), the number of three sides painted sliced cubes is always 8. The number of unpainted, double side painted and single side painted sliced cubes are given by formulae (X-2)^3, 12*(X-2) & (6X^2-24X+24). Only single side painted series is a bit complicated. The other two run like {0,1,8,27,64etc} and {0,12,24,36,48 etc}." Mohan

by:  N.MOHAN, TOYOTA, OMAN     nmohan@omantel.net.om  on 24 January 2008 		   
  January 20, 2008 The e4 of January 20, 2008 is an age old problem solved by 7th standard kids. In any cube of dimension “n” painted and cut uniformly, there are 8 cubes with 3 sides painted, 6(n-2)^2 cubes with one side painted, 12(n-2) cubes with two sides painted and the rest (n^3 – 8 – 6(n-2)^2 – 12(n-2)) cubes with no paint on any side. In fact one of the clues is enough.

by:  Ravi Kumar Meduri     ravi_meduri@satyam.com  on 24 January 2008 		   
  Here is the answer to Mindsport 20th January Cue Brute. The dimension of the cube is 6 units. For any such sliced cube, there would be a fixed number of sliced cubes with three-painted sides-namely 8. So for the cube in question, the number of sliced unpainted cubes is 64. Let us look at sliced cubes of dimension of X unit as side. There would be X stacks each containing X^2 cubes. The top and bottom stacks will have at least one side painted. Each of the remaining (X-2) stacks will contain the following distribution of painted cubes- 4 cubes of double side painted, 4(X-2) cubes of one side painted and hence the balance X^2-4-4(X-2) cubes of unpainted cubes. So if we solve the equation (X-2)* (X^2-4-4(X-2))=64, we find the answer as 6.

by:  N.MOHAN, TOYOTA, OMAN     nmohan@omantel.net.om  on 23 January 2008 		   
  Solution for the problem on cube painted red: First of all, the “dimensions” of the cube is incorrect; as a cube has only one “dimension”, its side. The cube in question has a side of 6 inches. The number of small cubes unpainted, one side painted and two sides painted, have standard relationships with the original cube’s size. (Three sides painted will always be 8 pieces, irrespective of the cube’s size.) For a cube of side n inches (n = 2 or more), the number of one-inch cubes unpainted will be (n-2)^3, one side painted 6*(n-2)^2, and two sides painted 12*(n-2). It may be noticed that while the coefficient is increasing in steps of 6, the exponent is decreasing gradually. According to condition (a) of the problem, 2*12*(n-2) = 6*(n-2)^2. This equation reduces to n-2 = 4, which further yields n = 6. So the condition (b) is absolutely redundant. Sheikh Sintha Mathar, sheikhsm@eim.ae

by:  Sheikh Sintha Mathar     sheikh@alrostamanigroup.ae  on 22 January 2008 		   
  Mr.Mukul , I am satisfying two of the conditions of running the responses.Firstly, the answer is correct & secondly I say "please,please". The answer to CUE BRUTE is that the dimensions of the cube will be 6"x6"x6". I got the answer following this procedure : Let the no.of cubes with two red sides be x. Thus, the no.of cubes with one red side will be 2x. Let the no. of cubes with three red sides be y. Thus, the no. of cubes without red sides will be 8y. Let the total no.of cubes be z. Now, we can write the above conditions in an equation form as z = x + 2x + y + 8y ie,z = 3x + 9y Now, we know that y = 8 SINCE ANY CUBE WILL HAVE 8 "THREE SIDES EXPOSED CORNERS" So, now we can say z = 3x + 72 Substituting z=27 or 64 gives a negative x,which is unacceptable.z=125 gives an undivisible x(also unacceptable) z=216 will yield x=48, which is perfect. So, the 6"x6"x6" cube will have 8 cubelets with three red sides 48 cubelets with two red sides 96 cubelets with one red side & 64 cubelets without any red side. making a total of 216 cubelets.

by:  Saifuddin Khomosi , Dubai     saif_sfk@hotmail.com  on 22 January 2008