# 400 puzzles

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- 1. 18Three friends divided some Find sum of digits of D.bullets equally. After all of themshot 4 bullets the total numberLetof bullets remaining is equal to A= 19991999the bullets each had after B = sum of digits of Adivision. Find the originalC = sum of digits of Bnumber divided.D = sum of digits of CAnswer (HINT : A = B = C = D (mod 9))18 AnswerAssume that initial there were The sum of the digits od D is3*X bullets. 1.So they got X bullets each after Let E = sum of digits of D.division. It follows from the hint that A =All of them shot 4 bullets. So E (mod 9)now they have (X - 4) bulletseach.But it is given that,after theyshot 4 bullets each, totalnumber of bullets remaining isequal to the bullets each hadafter division i.e. XTherefore, the equation is3 * (X - 4) = X3 * X - 12 = X2 * X = 12X=6Therefore the total bulletsbefore division is = 3 * X = 18 1

2. Consider,so 19991999 = 1 A = 19991999 (mod 9) < 20002000Therefore we conclude that = 22000 *E=1.10002000There is a 50m long army platoonmarching ahead. The last person in = 1024200 *the platoon wants to give a letter to106000the first person leading theplatoon. So while the platoon is < 10800 * 106000 marching he runs ahead, reachesthe first person and hands over the = 106800 letter to him and without stoppinghe runs and comes back to hisoriginal position. i.e. A < 106800In the mean time the wholeplatoon has moved ahead by 50m.i.e. B < 6800 *9 = 61200 The question is how much distancedid the last person cover in that i.e. C < 5 * 9 time. Assuming that he ran the= 45whole distance with uniformspeed. i.e. D < 2 * 9 Submitted= 18Answer i.e. E 1).I got this answer by writing1. On the first day 1 medal small program. If anyone know and 1/7 of the remaining any other simpler method, do m - 1 medals weresubmit it. awarded.2. On the second day 2A number of 9 digits has the medals and 1/7 of thefollowing properties: now remaining medals was awarded; and so on. The number comprising the leftmost two digits is divisible by 2, that7 8. comprising the leftmostcontainer evaporated on the 1stthree digits is divisible by day. 3/4th of the remaining3, the leftmost four by 4, contents of the containerthe leftmost five by 5, andevaporated on the second day.so on for the nine digits ofthe number i.e. theWhat part of the contents of thenumber formed from the container is left at the end offirst n digits is divisible by the second day?n, 2A, C->B5. A->C, C->B, B->ACASE 2 : King of Hearts is not6. A->C, C->B, B->Cdrawn from Pack A 7. A->C, C->A, B->A8. A->C, C->A, B->CProbability of not drawing Kingof Hearts from Pack A = 50/51(as Queen of Hearts is not to Out of which, there are only twobe drawn) cases under which the antsProbability of having King of wont collide :Hearts on the top of the Pack B= 1/53 A->B, B->C, C->A A->C, C->B, B->ASo total probability of case 2 =(50/51) * (1/53) = 50 / (51 * 53)Find all sets of consecutiveNow adding both the integers that add up to 1000.probability, the required Submitted by : Jamesprobability isBarberousse= 2 / (51 * 53) + 50 / (51 * 53)= 52 / (51 * 53)= 52 / 2703Answer= 0.0192378There are 3 ants at 3 corners of There are total 8 such series:a triangle, they randomly startmoving towards another corner.1. Sum of 2000 numbers starting from -999 i.e.What is the probability that theysummation of numbersdont collide? from -999 to 1000. (-999) + (-998) + (-997)Answer + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 +Lets mark the corners of the999 + 1000 = 1000triangle as A,B,C. There are2. Sum of 400 numberstotal 8 ways in which ants can starting from -197 i.e.move.summation of numbers from -197 to 202. 1. A->B, B->C, C->A (-197) + (-196) + (-195) 2. A->B, B->C, C->B + ..... + (-1) + 0 + 1 + 2 3. A->B, B->A, C->A24 25. + ..... + 199 + 200 + summation of numbers 201 + 202 = 1000from 198 to 202.3. Sum of 125 numbers198 + 199 + 200 +201 starting from -54 i.e.+ 202 = 1000 summation of numbers 8. Sum of 1 number from -54 to 70. starting from 1000. (-54) + (-53) + (-52) + 1000 = 1000 ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 10004. Sum of 80 numbers starting from -27 i.e. There is a 4-character code, summation of numbers with 2 of them being letters and from -27 to 52.the other 2 being numbers. (-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + How many maximum attempts ..... + 50 + 51 + 52 = would be necessary to find the 1000 correct code? Note that the5. Sum of 25 numberscode is case-sensitive. starting from 28 i.e.Answer summation of numbers from 28 to 52. The maximum number of 28 + 29 + 30 + 31 + 32 attempts required are + 33 + 34 + 35 + 36 +16,22,400 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 +There are 52 possible letters - a 46 + 47 + 48 + 49 + 50 to z and A to Z, and 10 possible + 51 + 52 = 1000 numbers - 0 to 9. Now, 46. Sum of 16 numberscharacters - 2 letters and 2 starting from 55 i.e.numbers, can be selected in summation of numbers 52*52*10*10 ways. These 4 from 55 to 70. characters can be arranged in4 55 + 56 + 57 + 58 + 59 C2 i.e. 6 different ways - the +60 + 61 + 62 + 63 + number of unique patterns that 64 + 65 + 66 + 67 + 68 can be formed by lining up 4 + 69 + 70 = 1000 objects of which 2 are7. Sum of 5 numbers distinguished one way (i.e. they starting from 198 i.e. must be letters) and the other 2are distinguished another way 25 26. (i.e. they must be numbers). first.Consider an example : LetsThere are 8 corner cubes,assume that @ represents which can be arranged in 8!letter and # represents number.ways.the 6 possible ways of Each of these 8 cubes can bearranging them are : @@##, turned in 3 different directions,@#@#, @##@, #@@#,so there are 3^8 orientations#@#@, ##@@ altogether. But if you get all but one of the corner cube intoHence, the required answer ischosen positions and= 52*52*10*10*6orientations, only one of 3= 16,22,400 attempts orientations of the final corner= 1.6 million approx.cube is possible. Thus, total ways corner cubes can beThanks to Tim Sanders forplaced = (8!) * (3^8)/8 = (8!) *opening BrainVistas brain !!! (3^7)How many possiblecombinations are there in aSimilarly, 12 edge cubes can3x3x3 rubics cube? be arranged in 12! ways. Each of these 12 cubes can beIn other words, if you wanted to turned in 2 different directions,solve the rubics cube by tryingso there are 2^12 orientationsdifferent combinations, howaltogether. But if you get all butmany might it take you (worstone of the edge cube intocase senerio)? chosen positions and orientations, only one of 2How many for a 4x4x4 cube? orientations of the final edgeSubmittedcube is possible. Thus, total ways edge cubes can beAnswer placed = (12!) * (2^12)/2 = (12!) * (2^11)There are 4.3252 * 10^19possible combinations forHere, we have essentially3x3x3 Rubics and 7.4012 *pulled the cubes apart and10^45 possible combinationsstuck cubes back in placefor 4x4x4 Rubics.wherever we please. In reality, we can only move cubes around by turning the faces ofLets consider 3x3x3 Rubicsthe cubes. It turns out that you 26 27. cant turn the faces in such aN E Vway as to switch the positionsE Rof two cubes while returning allthe others to their originalL E Apositions. Thus if you get all butV Etwo cubes in place, there isonly one attainable choice for +them (not 2!). Hence, we must M Edivide by 2. -----------Total different possible------combinations are= [(8!) * (3^7)] * [(12!) * (2^11)] / A L O2 N E= (8!) * (3^7) * (12!) * (2^10) Note that the leftmost letter= 4.3252 * 10^19cant be zero in any word. Also,there must be a one-to-onemapping between digits andSimilarly, for 4x4x4 Rubics total letters. e.g. if you substitute 3different possible combinations for the letter M, no other letterare can be 3 and all other M in the= [(8!) * (3^7)] * [(24!)] * [(24!) / puzzle must be 3.(4!^6)] / 24= 7.4011968 * 10^45 AnswerNote that there are 24 edge A tough one!!!cubes, which you can not turnin 2 orientations (hence no Since R + E + E = 10 + E, it is2^24 / 2). Also, there are 4clear that R + E = 10 andcenter cubes per face i.e. (24!) /neither R nor E is equal to 0 or(4!^6). You can switch 2 cubes5. This is the only entry point towithout affecting the rest of thecombination as 4*4*4 has even solve it. Now use trial-n-errordimensions (hence no division method.by 2). But pattern on one side isrotated in 4 directions over 6faces, hence divide by 24.Substitute digits for the lettersto make the following relationtrue.27 28. sex then the man is olderN E V E R than the woman.21 4 1 9 5. If the dancer is a woman,then the dancer is olderL E A V E than the singer.31 5 4 1Whose occupation do you+M Eknow? And what is his/her+6 1occupation?Answer---------------------------------- Cindy is the Singer. Mr.Clinton or Monika is theA L O N E Dancer.53 0 2 1From (1) and (3), the singerand the dancer, both can not beOne of the four people - Mr.a man. From (3) and (4), if theClinton, his wife Monika, their singer is a man, then theson Mandy and their daughterdancer must be a man. Hence,Cindy - is a singer and another the singer must be a woman.is a dancer. Mr. Clinton is olderthan his wife and Mady is older CASE I : Singer is a womanthan his sister.and Dancer is also a womanThen, the dancer is Monika and1. If the singer and thethe singer is Cindy. dancer are the same sex, then the dancer is older CASE II : Singer is a woman than the singer. and Dancer is also a man2. If neither the singer norThen, the dancer is Mr. Clinton the dancer is the parent and the singer is Cindy. of the other, then the singer is older than the In both the cases, we know that dancer.Cindy is the Singer. And either3. If the singer is a man,Mr. Clinton or Monika is the then the singer and theDancer. dancer are the same age. There are 20 people in your4. If the singer and theapplicant pool, including 5 pairs dancer are of opposite of identical twins. 28 29. If you hire 5 people randomly,10C3 * what a