Question

$\frac{{32}^{{32}^{32}}}{9}$ will leave a remainder

• A. 4
• B. 7
• C. 1
• D. 2

Answer 1

The correct option is A 4

${32}^{{32}^{32}}÷9\to 5^{{32}^{32}}÷9=5^{6n+x}÷9$. We write this in the form of $5^{6n+x}$ because $5^6$ leaves a remainder of 1 when divided by 9. When we try to see ${32}^{32}$ as 6n +x, we can find the value of x as the remainder of $2^{32}$ when divided by 6. The following thought process would help us find this value:
$2^{32}÷6=2^{31}÷3\to$ Remanider =2 (by the $a^n÷(a+1)$ rule). Thus, $2^{32}÷6$ would have a remainder of $2\times 2=4.$
Hence, the required remainder would be $5^4÷9$, which is 4.