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Question

If aabb is a 4 digit number and also a perfect square then the value of a+b is?

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Solution

Number aabb can be written in expanded from as,

aabb = 1000a + 100a + 10b + b = 1100a + 11b = 11(100a + b)

For aabb to be a perfect square, 100a + b should be of the form 11n^2

, where n is a natural number.

aabb = 11 × 11 × n^2

When n = 4,

11 × 11 × n^2

= 121 × 16 = 1936. This is not in the form aabb.

When n = 5,

11 × 11 × n^2

= 121 × 25 = 3025. This is not in the form aabb.

When n = 6,

11 × 11 × n^2

= 121 × 36 = 4356. This is not in the form aabb.

when n = 7,

11 × 11 × n^2

= 121 × 49 = 5629. This is not in the form aabb.

When n = 8,

11 × 11 × n^2

= 121 × 64 = 7744. This is in the form aabb.

When n = 9,

11 × 11 × n^2

= 121 × 81 = 9801. This is not in the form aabb.

So, 7744 is four digit number.

a + b = 7 + 4 = 11


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