By Maxey Brooke
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Extra info for 150 Puzzles in Crypt-Arithmetic
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. (d) We next prove that E = 7 since E is greater than A or C. E is not 1, 2, or 3 (the fourth product contains six digits), nor is it 5 or 6 because in those cases D = E. Therefore E = 4, 7, 8, or 9. The only value leading to a solution is 7.
One multiple of8: 1384. One multiple of 16: 3184. 30. The square of C ends in C, thus C must be 0, 1, 5, or 6. Since ABC is prime, C = 1. For the square BC to end in BC, B must equal zero. ABC can be 101,401,601, or 701. And only 601 is prime. 31. C = 0 and G = 1. J = H + I, A = 2, B = 4. Thus: 240)893: 720: 173: 168: 5: 4: 32. Let 1000·0+ 100·D+ lO·E+R = 18(1O·D+0+1O·D+R).
The result was that in the October 27, 1955, issue of the Toronto Globe & Mail there appeared FUN WITH FIGURES by J. A. H. Hunter It's just an easy alphametic today: merely a matter of commonsense and a little patience. Each letter stands for a different figure, and you're asked to find the number represented by ABLE. BE)ABLE(SIR MR RRL RLM BE BE So a new name was born. The following eight puzzles are from Mr. Hunter's fertile brain. 46 PUZZLES 145. NEVER -DRIVE RIDE 146. FIT)POLE (TO LOT FREE FREE 147.