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