By Maxey Brooke
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Extra info for 150 Puzzles in Crypt-Arithmetic
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.
D) Since D x B has only one digit, B must be 3 or less. The only two possibilities are 0 and 2. B cannot be zero because 7B is a two-digit number. Thus B = 2. (e) By completing the multiplication, F = 8, E = 7, and G = 6. (f) The answer is 125 x 37 = 4625. (a) T must equal I and W ~ 4, since THREE contains five digits. (b) E = 4 because 44 is the only combination of two equal digits that terminate a square. Hence 0 = 2 or 8. (c) The answer is TWO = 138. 2. 3. There are 22 ways of satisfying the relation I x RL = TS if we make the obvious restrictions I # 1,5,9 and L, S # 0, 1,5.
Since the remainder of the division is zero, the product I x I terminates in E, a digit different from I, which means I cannot equal 0, 1, 5, or 6. The product QUI x C terminates in E. Thus C and I, two different digits, when multiplied by I, give products ending in E. 54 ANSWERS Therefore I cannot be 1, 3, 7, or 9. This leaves only three values for I, each of which establishes values for E and C. I E C 843 469 2 4 7 CECI 343 8 9 694 74 72 The first hypothesis requires that QUI x 8 be a number of three digits, so QUI < 125.
150 Puzzles in Crypt-Arithmetic by Maxey Brooke