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.
Algebra ->
Customizable Word Problem Solvers
-> Numbers
-> 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.
Log On
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
