Determine the (non-leading zero) 3-digit positive integer N such that the product of the digits of N is equal to N/5.
Source: Adapted from a problem appearing in a German Mathematical Olympiad.
*** Computer solutions are welcome, but an analytical solution is preferred.
Call the 3 digits of N, from left to right, a, b, and c.
Since we are given N/5 = a*b*c, N must be divisible by 5. So c must equal either 0 or 5. However, c = 0 would imply that N/5 = a*b*c = 0, which violates the condition that N has no leading zeros. So c = 5.
Now, since N/5 = 5*a*b, N = 25*a*b. Since N is a multiple of 25 and c = 5, b must equal 2 or 7. However, if b = 2, the product a*b*c is a multiple of 10, which implies that N = 5*a*b*c is also a multiple of 10. Since we have already determined that c = 5, this is a contradiction, so b = 7.
We now have a single equation and a single unknown:
[100a + 75]/5 = a*7*5
20a + 15 = 35a
a = 1
So the unique solution is N = 175.
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Posted by H M
on 2022-10-04 07:43:03 |
Analytical Solution
