Numbers... without numbers 3

Consider a two digit number AB

Where:

A+B = C

C x C = A B \overline{AB}

What is A B \overline { AB } ?

Details and assumptions:

Each letter represents one numerical digit. AB does not indicate multiplication. AB is positive.


The answer is 81.

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3 solutions

Alvi Newaz
Jul 31, 2014

C*C=AB so 9<C^2<100. So just find the squares of 4,5,6,7,8,9 and one of these is the solution.

Dan Wilhelm
Jul 7, 2015

It is given that the number A B \overline{AB} is two digits and a perfect square.

There are only six numbers that are two-digit perfect squares: 16 , 25 , 36 , 49 , 64 , 81 16, 25, 36, 49, 64, 81 .

From the problem statement, A B \overline{AB} must also satisfy the condition: ( A + B ) 2 = A B (A + B)^2 = \overline{AB} .

Testing each of the six perfect squares, only 81 81 satisfies this condition: ( 8 + 1 ) 2 = 81 (8 + 1)^2 = 81 .

Hence, A B = 81 \overline{AB} = 81 .

Because AB is a 2-digit number:
9 < A B = C 2 = ( A + B ) 2 < 100 9<\overline { AB } ={ C }^{ 2 }={ (A+B) }^{ 2 }<100
So AB must be 16, 25, 36, 49, 64 or 81 (squares of 4,5,6,7,8,9). And we have 3 < A + B < 10 3<A+B<10
So we get rid of 49 and 64, as their A+B > 10.
( A + B ) 2 = ( 8 + 1 ) 2 = 9 2 = 81 { (A+B) }^{ 2 }={ (8+1) }^{ 2 }={ 9 }^{ 2 }=81
So it's 81.
(I know it looks like Alvi Newaz's solution, I just want to make it clearer and somehow easier)




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