Download e-book for kindle: 150 Puzzles in Crypt-Arithmetic by Maxey Brooke

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Extra resources for 150 Puzzles in Crypt-Arithmetic

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C) VALSE x 2 is smaller than 100,000, making V = 3 and leaving A = 7. (d) 37215 x 12735. 24. 25. The last partial remainder ends in 000 and the next-to-Iast remainder ends in o. The divisor does not end in zero, otherwise the next-to-Iast partial product would end in zero and the rest would be zero. l a multiple of 5, the next-to-Iast final product ends in 5 and the last partial product is 5000. 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.

B) Since A x C ends in A and B x C ends in B, C must be either 1 or 6. But since the first product contains four digits, C > 1, hence C = 6. (c) A and B are even (but not 6), hence A and Bare 2, 4, or 8. Now, A cannot be 4 or 8 since the second product contains three digits, hence A is 2. (d) B must be 4 or 8. B cannot be 4 because the last product contains four digits. Hence B is 8. (e) The multiplication is 286 x 826. 6. (a) The square ofD does not end in D, hence D cannot be 0, 1, 5, or 6. C must equal 1, 4, or 9.

The condition Ex BTO = RTMP gives the unique solution: 5 11960 4 1392 7 198 14 So: 0 P 1 R 2 0 3 B 4 L 51 5 6 E M 7 I 8 S 9 T 52 4. ANSWERS This is Berwick's solution of his own problem: (a) Since the multiplication of the divisor by 7 gives 6 figures in the product, while the second and fourth multiplications give 7 figures, the divisor must begin with 11, 12, 13, or 14, and the second and fourth figures of the quotient are 8 or 9. (b) Since the product of the divisor multiplied by 7 has as its second digit 7, it is found by trial that the divisor must begin with 111, 124, 125, 138, or 139.

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150 Puzzles in Crypt-Arithmetic by Maxey Brooke


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