2015/048) What could be possible value of integer a if $10^ {2n +1} + a.7^ {2n +1}$ is divisible by 51 exactly

we have $10^2 = 100 = -2$ mod 51

so $10^ {2n +1} = 10(-2)^n$

$7^2 = -2$ mod 51

so $a.7^ {2n +1} = 7a (-2)^n$ mod 51

$10^{2n +1} + a.7^ {2n +1}$ mod 51
= $10(-2)^n  + 7a (-2)^n = 0$ as divisible by 51

so $10 + 7a = 0$ mod 51
$7a = - 10$ mod 51 = 41 mod 51

$7a = 41$ mod 51

we need to find inverse of 7 mod 51( we can find by extended euclid algorithm as below)

$51 = 7 * 7 + 2$  or  $2 = 51- 7 * 7$

$7 = 2 * 3 + 1$ or $1= 7 - 2 * 3 = 7 - (51- 7 * 7) * 3 = 51 * 3 - 22 * 7$

so 22 = inverse of 7 mod 51

so $a = 41 * 22$ mod 51 or 35 mod 51 or 51k + 35