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420
Represent the digits of the number as a, b, and c. Thus, the number itself is equal to 100a + 10b + c.
The statements in the problem give the following equations:
a² + b² + c² = 10b + c
(10b + c)(10b + c) + a² + a = 100a + 10b + c
With some manipulation we can combine the equations:
100b² + 20bc + c² + a² + a = 100a + a² + b² + c²
99b² + 20bc = 99a
99(b² - a) + 20bc = 0
This last equation leads to one of two conclusions.
First, it could be that (b² - a) is negative and 20bc is a multiple of 99. The lowest common multiple of 99 and 20 is 990, which would imply that (b² - a) were -10. Since a and b are both positive, single-digit integers, that is not possible.
So, it must be that b² - a = 0 and bc = 0. A trivial case would be where a and b are both 0; however, the digits must all be different. Otherwise, c must be zero, and a = b².
For single digits, there are only three possibilities for one number being the square of the other: 1 and 1, 2 and 4, or 3 and 9 (110, 420, or 930). 110 is immediately ruled out because all the digits must be different.
Some of the information from the original first statement was lost in the combination, so we need to check these numbers against it again. (4² + 2² + 0²) is 20, and (9² + 3² + 0²) is 33, not 30. So, it appears that the answer is 420.
Just to be sure, we can recheck this number against the second statement as well, and see that (20² + 4² + 4) does indeed equal 420. |
