13197044 Brain Vista Puzzles

PUZZLES FROM BRAINVISTA Compiled by Raji

Sam and Appu are intelligent mathematicians. Both of them have two prime numbers written on their foreheads and are told that any three of them form the sides of a triangle with prime perimeter. They take turns stating whether they can deduce the numbers on their own foreheads. Sam sees 13 and 17 on Appu's forehead and each of them has stated "Don't know" on each of their first two turns. It is now Sam's third turn. What are the numbers on Sam's forehead?

Answer The numbers on Sam's forhead are 13 and 17. Corresponding to 13 and 17 on Appu's head, there are just 3 combinations which satisfy prime perimeter triangle condition: (7, 17), (11, 13) and (13, 17) If Sam had 7 and 17, Appu would have guessed 13 and 17 in his first guess as that is the only possible solution. If Sam had 11 and 13, Appu had one of the 2 choices to make (5, 13) and (13, 17). But Appu being an intelligent mathematician would never had chosen (5, 13) because if that was the case, his counterpart would have guessed (11, 13) in his first guess. But since Sam passed on his first attempt, Appu knows it cannot be (5, 13). Incidentally Appu is not able to guess it in the second turn also so even (11, 13) as an option is ruled out. The last option is (13, 17) which like before has 3 possibilities. Since both pass first two times only that option remains and Sam guesses it right in his third attempt. Hence, the numbers on Sam's forhead are (13, 17).

Santa and Banta are in charge of counting the people who get on and off the elevator in a hotel. They take turns riding to the top floor and back down, counting as they go. After two such trips each morning, two around noon and two in the evening, there is an average taken. The hotel manager wants to know today's average. 1. On Banta's noon trip there were 32 fewer people than in his morning count. 2. Santa counted a total of 122 in the morning and noon counts, just one higher than his evening count, but 24 more than Banta's evening count. 3. Santa's morning count is the same as Banta's average. 4. Santa's evening count was 37 more than Banta's morning count.

Answer From (2), Santa's evening count was 121 and Newell's evening count was 98. From (4), Santa's evening count was 37 more than Newell's morning count. Hence, Newell's morning count was 84. From (1), Banta's noon count was 32 less than his morning count. Hence, Banta's noon count was 52. Thus, Banta's total count = (84 + 52 + 98) = 234 Average = 234 / 3 = 78 From (3), Santa's morning count was same as Banta's average. Hence, Santa's morning count was 78. From (2), Santa's total count for morning and noon was 122. Hence, Santa's noon count was 44 (122 - 78). Thus, Santa's total count = (78 + 44 + 121) = 243 Average = 243 / 3 = 81

Morning Noon Evening Total Average Santa

78

44

121

243

81

Banta

84

52

98

234

78

Substitute digits for the letters to make the following addition true. Y E A H T H A T S +

O U R

------------------S T O R Y Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 5 for the letter T, no other letter can be 5 and all other T in the puzzle must be 5.

Answer A=1, T=2, S=3, U=4, H=5, Y=6, O=7, R=8, E=9 Following can be deduced from the given cryptogram: 1) S = T + 1 2) E + A = 9 or 10 3) Y + H >= 10 4) 1 OX i.e. smaller sphere requires more space than the space available. Hence, smaller sphere of 7 cms diameter can not pass through the space between the big sphere, the wall and the floor. The puzzle can be solved by another method. Draw a line tangent to the big sphere at the point X such that X is the closest point to the origin O on sphere. The tanget will cut X and Y axes at A and B respectively such that OA=OB. [See Fig III] From above, OX=8.28427 cms. From the right angle triangle OAB, we can deduct that OA = OB = 11.71572 cms AB = 16.56854 cms

Now, the diameter of the inscribed circle of right angle triangle is given by d = a + b - c where a