Problem 150 Solution

It is obvious that A can be no more than 2. If A were 3 then 3BCDE * 4 would be at least 120,000 which is more than five digits. Also A must be an even number because EDCBA is an even number since it is the product of at least one even number (4). We can eliminate A=0 because E would have to be 5 (5*4=0) but BCDE*4 could not hope to reach 50,000. So A must be 2.

Next consider E. E*4 must end in the digit 2. The only numbers that works for are 3 and 8. However with A=2 EDCBA must be at least 80,000. So 8 is the only number that satisifies both conditions.

Next consider B. We already know that 2BCD8*4 is at least 80000 and less than 90000. B can not be more than 2 because then 2BCD8 * 4 would be more than 80000. 2 is already taken so B must be 0 or 1. Lets consider the case that B=0. Then D8 * 4 must end in the digit 02. However there is no D that satisfies this condition. So B must be 1.

Next consider D. D8*4 must end in the digits 12. The only possiblity is D=7 (78*4=312).

Now solve for C:

21C78 * 4 = 87C12.
84312+400C = 87012 + 100C
2700 = 300C
C=2700/300=9.

So ABCDE=21978.

I'd like to thank "Marisa" for this problem.

Michael Shackleford, ASA, December 28, 1999

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