By Maxey Brooke
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Extra info for 150 Puzzles in Crypt-Arithmetic
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. 20. R must equal zero. 2U + F ;;:; 9, hence U = 1, 2, 3, or 4. And USA = 178, FDR = 230, and NRA = 408. 21. (a) B x D and C x D give two-digit numbers ending in D. This is possible only ifD = 5. Band C must equal 3, 7, or 9. (b) If B = 3, *D = 15; if B = 7, *D = 35; if B = 9, *D = 45.
AAA x BC = ADADA BBB x CA = CCBEE PUZZLES 26 65. ABC C A DEB BED A C C B A By replacing the letters with numbers, and reading right to left, up and down, left to right, and down and up, you obtain eight prime numbers, each used twice. To save a long research in a table of primes, it is necessary only to know that the sum of the digits of the four numbers forming the sides of the square is 60, and the sum of the digits of the four numbers forming the interior square is 16. 66. QUI PREND *** **** *** *** CA ) * * * * * * * ( * * * * * * ** *** *** *** *** o 67.
DID) JUDGE (IT x x x x IDLE IDLE PUZZLES 47 150. TIE IT SAVE SPAR STEVE 151. PENITENT - SPINSTER REPINED 152. WAS) MA SK ED (W AS xxxx xxxx xxxx xxxx Axxx SPY 153. Here is a new form of cryptarithm. Each digit can be represented by two letters, but each letter represents only one digit. VDOUBLE(SIR X H I ILET PU B HYR IGALE GALE 48 PUZZLES 154. This problem started life as a pictorial cryptarithm, but ended up like this: BE)ABLE (MAN MN LL AT HE HE 155. YOU P a a a a a a a THaNK 156. And finally, an anti-cryptarithm: Prove NA· CL =f.