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Question 962664: A certain number has 2 digits and is equal to the square of the sum of the digits. If the number is decreased by 63, the digits are interchanged. Find out the original numbers
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! a = the 10's digit
b = the units
ten
10a+b = "the number"
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Write an equation for each statement, simplify as much as possible
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A certain number has 2 digits and is equal to the square of the sum of the digits.
10a + b = (a+b)^2
let's leave it at that, see if we can get it down to a single unknown
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If the number is decreased by 63, the digits are interchanged.
10a + b - 63 = 10b + a
combine on the left
10a - a + b - 10b = 63
9a - 9b = 63
simplify, divide by 9
a - b = 7
a = (b+7)
Back to the 1st equation, replace a with (b+7)
10(b+7) + b = (b+7+b)^2
10b + 70 + b = (2b+7)^2
11b + 70 = 4b^2 + 28b + 49
Combine to form a quadratic equation on the right
0 = 4b^2 + 28b - 11b + 49 - 70
4b^2 + 17b - 21 = 0
you can use the quadratic formula, but this has to factor (integers)
(b-1)(4b+21) = 0
b = 1, the only reasonable solution (has to be an integer)
then
a = 1 + 7
a = 8
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Find out the original numbers, 81 is the original number
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See for yourself if it checks out in the original statements.
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