What is the value of x (0 \leq x \leq 8) if (100^{97}+100^{54}+x+1) leaves a remainder 0 when divided by 9?

Explanation

When dividing by 9, 100 \equiv 1 \pmod 9. Therefore, 100^{97} + 100^{54} + x + 1 \equiv 1^{97} + 1^{54} + x + 1 \pmod 9. Simplifying gives 1 + 1 + x + 1 = x + 3. For the number to be perfectly divisible by 9, x + 3 = 9 \implies x = 6.

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