Solve the above equation for A,B,C,D, and E where each is a unique integer from 0 to 9.
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. Let’s 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).
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