SOLUTION: Find a four-digit perfect square the sume of whose digits is 22 and which has its first two digits equal to each other and its last two digits equal to each other.
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Question 538468: Find a four-digit perfect square the sume of whose digits is 22 and which has its first two digits equal to each other and its last two digits equal to each other. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Find a four-digit perfect square, the sum of whose digits is 22 and
which has its first two digits equal to each other, and it's last two digits equal to each other.
:
Let x = the first two digits
Let y = the last two digits
:
2x + 2y = 22
simplify, divide by 2
x + y = 11
:
Try and see method, I came up with 7744: = 88
