# The five digit number 2a9b1 is a perfect square. What is the value of a^(b-1)+b^(a-1)?

Feb 9, 2017

$21$

#### Explanation:

As $2 a 9 b 1$ is a five digit number and perfect square, the number is a $3$ digit number and as unit digit is $1$ in the square, in square root, we have either $1$ or $9$ as units digit (as other digits will not make unit digit $1$).

Further as first digit in square $2 a 9 b 1$, in the place of ten-thousand is $2$, we must have $1$ in hundreds 'place in square root. Further as first three digits are $2 a 9$ and $\sqrt{209} > 14$ and $\sqrt{299} \le 17$.

Hence, numbers can only be $149$, $151$, $159$, $161$, $169$, $171$ as for $141$ and $179$, squares will have $1$ or $3$ in ten thousands place.

Of these only ${161}^{2} = 25921$ falls as per pattern $2 a 9 b 1$ and hence $a = 5$ and $b = 2$ and hence

${a}^{b - 1} + {b}^{a - 1} = {5}^{2 - 1} + {2}^{5 - 1} = {5}^{1} + {2}^{4} = 5 + 16 = 21$