An algebra problem by alok poonia
Algebra
Level
2
abcde is a five digit no such that
abcde × 4 = edcba
Then find the value of a+b+c+d+e
The answer is 27.
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1 solution
First, it is obvious that A must be an even number, because we are multiplying by 4 (an even number). The last digit will therefore be even. It can't be 0, because that would make ABCDE a four-digit number. It can't be more than 2, because that would result in a six-digit answer. So A = 2.
2BCDE x 4 = EDCB2
So what can E be? The choices are E = {3, 8} because 3 x 4 = 12 and 8 x 4 = 32. But a value of 3 doesn't work in the result (3????) because it is too small.
2BCD8 x 4 = 8DCB2
Since the final number is 8 and we have 2 x 4, that means there is no carry from the prior multiplication (4 x B + carry). So B can't be anything higher than 1, possibly 0.
Looking at the other side of the equation, we have 4D + 3 = (a number ending in 0 or 1). In other words, 4D must end in 7 or 8. Obviously only 8 works, because 4 is an even number. Working forward again, that means B = 1.
21CD8 x 4 = 8DC12
So what values of 4D result in a number ending 8? 4 x 2 = 8, 4 x 7 = 28. Now 2 is already taken and the problem said the digits were unique. So D = 7.
21C78 x 4 = 87C12
Finally, we have a carry of 3 (from 28 + 3 = 31). And when we calculate 4C + 3 it must also result in a carry of 3 and a last digit of C. In other words: 4C + 3 = 30 + C
This is easy to solve: 3C = 27 C = 9
Thus the final answer is: 21978 x 4 = 87912