Download 150 Puzzles in Crypt-Arithmetic by Maxey Brooke PDF

By Maxey Brooke

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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. (d) 5+ T + 1 = R, a two-digit number. So T = 4, 6, or 8. R > 2 and since E + R = E, R = 0 and T = 4. (e) P = 1. (f) SEPTEMBRE = 721425902.

3 3 ... . 33 80. GALON and ALONG are the squares of GOO and OOG to the base 6. 30 PUZZLES 81. When the numbers 1234567890 are substituted for an English word, it is possible to make the following magic square. The sums of the rows, columns, and diagonals are 275. RL AA CN AE UA BR BN MC ML MB EE EU LM LD NC BD EA LR LN UU RC RB AU UE BM 82. There are eight brothers and sisters in a family: Georgette, Ulysse, Yvon, Marie, Annette, Roger, Isadore, and Emile. All are less than 10 years old. If you represent the age of each child by the first letter of its name, you obtain YU GUY )MAR IE MARE EE If Marie is the youngest of the sisters, what is her age?

B) CD x D terminates with DC, and the only groups meeting this condition are 97 or 48. D x B terminates in C. Assume CD = 97; then B = 7 = D, which is impossible. (c) Thus CD = 48, which means B must equal 3 or 8. But since D = 8, B must equal 3 and BCD = 348. Since BCD x D terminates in ADC, A must equal 7, and ABCD = 7348. 7. (a) None of the letters A, B, C, D, or E are zero. (b) A and C are smaller than B, D, or E because their products contains only five digits. (c) A < 3 because if A = 3, B is at least 4 and 3 x 34000 is greater than 100,000.

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