SOLUTION: The 1st, 2nd, and 3rd digits are all the same. Together they make a three-digit number that is 70 less than the square of the product of the first two digits.

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Question 784154: The 1st, 2nd, and 3rd digits are all the same. Together they make a three-digit number that is 70 less than the square of the product of the first two digits.
Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
The 1st, 2nd, and 3rd digits are all the same.
100n+10n+n = 11n
Together they make a three-digit number that is 70 less than the square of the product of the first two digits.
111n = (n*n)^2 - 70
111n = (n^2)^2 - 70
111n = n^4 - 70
n^4 - 111n - 70 = 0
Find the positive solution for this equation by graphing

n = 5
:
:
See if that works in the original statement
555 = (5*5)^2 - 70
555 = 625 - 70

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