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# MATHCOUNTS Problem

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There is at least one two-digit number such that when it is added to the two-digit number having the same digits in reverse order, the sum is a perfect square. Find the sum of all such two-digit numbers.

benjamingu22  Jul 20, 2017
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#1
+26484
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29 + 92 = 121

38 + 83 = 121

47 + 74 = 121

56 + 65 = 121

.

Alan  Jul 20, 2017
#2
+82686
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Here's how Alan may have found his answers....

[And then again, maybe not !!! ]

Let the first number be  10 a +  b

And let the other number be  10 b + a

And we know that

[10 a +  b] + [10 b + a]  = N^2

10a + a + 10b + b  = N^2

11a + 11b = N^2

11 ( a + b)  =  N^2

But.....  if the right side is a perfect square then ( a + b)  must equal 11

So.....the possibilities are

9,2

8,3

7,4

6,5

And subbing these into 10a + b  and 10b + a  will generate Alan's two digit numbers

CPhill  Jul 20, 2017

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