By Maxey Brooke
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Additional resources for 150 Puzzles in Crypt-Arithmetic
So T = 5 and E is even (2 or 4). Since Ex E***T = T****, E must equal 2. (b) The left-hand digit of the multiplier is 1. (c) P = 4. (d) All the T's are given, so the digit on the right of the third partial product is zero, and the second digit from the left of the multiplier is 8. (e) The multiplier is 1825 = 52 x 73. (f) Because PIGEOLET is a cube, it is the cube of 5 x 73. 19. (a) lO+E-S = M S-E = M 2M = 10 M = 5 (b) 2E+ 1 ends in 5, so E is either 2 or 7. S-E = 5, so E = 2 and S = 7. (c) B = 9.
The last digit of the quotient is 8 and the divisor is 625. 1008. ANSWERS 59 26. The difference between the squares of two consecutive numbers is equal to the sum of the numbers, hence: OMLI - OHEM = LET But RG + RA = LET. Hence L = 1. The subtraction gives E = 5 or zero. If E = 5, R = 7, which is impossible since the sum of the squares of two consecutive numbers cannot end in 1, 3, or 5. Thus E = zero and R = 5 while 0 = 2. The sum of A and G is less than 10, so G = 3 and A = 4. The key is LOGARITHME.
BDAC must be either 9317 or 3917 as Band D equal 9 and 3. But 9317 is divisible by 7 whereas 3917 is prime. Thus A = 1, B = 3, C = 7, and D = 9. 29. (a) The four digits are all different, else there would not be 24 different permutations. (b) Since there are 24 permutations, 12 odd and 12 even, then two digits are odd and two are even. 60 ANSWERS (c) One of the odd digits must be 1 or 9 because the square of a prime must end in one of these numbers; however, the odd digit cannot be 5 because a prime cannot end in 5.
150 Puzzles in Crypt-Arithmetic by Maxey Brooke
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