1993 AJHSME Problems/Problem 20

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Problem

When $10^{93}-93$ is expressed as a single whole number, the sum of the digits is

$\text{(A)}\ 10 \qquad \text{(B)}\ 93 \qquad \text{(C)}\ 819 \qquad \text{(D)}\ 826 \qquad \text{(E)}\ 833$

Solution

\begin{align*} 10^2-93&=7\\ 10^3-93&=907\\ 10^4-93&=9907\\ \end{align*}

This can be generalized into $10^n-93$ is equal is $n-2$ nines followed by the digits $07$. Then $10^{93}-93$ is equal to $91$ nines followed by $07$. The sum of the digits is equal to $9(91)+7=819+7=\boxed{\text{(D)}\ 826}$.

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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