**kenig** (07/16/82)

5 4 3 2 1 The problem: ABCDE x 4 ----- EDCBA Analysis: Step 1: Realize that A must be either 1 or 2. REASON: obvious since 4 x A <10 (or the answer would be nEDCBA due to a carry). IF A = 1 then There must exist an E such that: 1) E = 4 and, 2) 4 x E = n1 for some digit n A contradiction since we all learned in 1st grade that 4x4=16. A = 2 -> E = 8 E x 4 = 8 x 4 = 32 (consistant since A is digit 2). RESTATEMENT of partially solved problem: 2BCD8 x 4 ----- 8DCB2 Step 2: Assert 4 x B < 10 or a carry would add to 4 x A and make the problem impossible. Since a already equals 2, B must be 1. 21CD8 x 4 ----- 8DC12 Step 3: Find 3 + (4 x D) = n1 for some n. By simple elimination, the answer is 7. 3 + (4 x D) = 31 21C78 x 4 ----- 87C12 Step 4: Assert 4 x C > 30 since a 3 must be carried to the 4 x B (4 x 1) column. either C = 8 or C = 9. Since 8 is already equal to E, C must equal 9. 21978 x 4 ----- 87912 Q. E. D. if not a bit too formally for such trivia. Marc Kenig BTL (...houxf!kenig) P.S. how about more toughies like the pizza problem, please?