kenig (07/16/82)
5
4
3
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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?