2007 AMC 10B Problems/Problem 24

Problem

Let $n$ denote the smallest positive integer that is divisible by both $4$ and $9,$ and whose base-$10$ representation consists of only $4$'s and $9$'s, with at least one of each. What are the last four digits of $n?$

$\textbf{(A) } 4444 \qquad\textbf{(B) } 4494 \qquad\textbf{(C) } 4944 \qquad\textbf{(D) } 9444 \qquad\textbf{(E) } 9944$

Solution

For a number to be divisible by $4,$ the last two digits have to be divisible by $4.$ That means the last two digits of this integer must be $4.$

For a number to be divisible by $9,$ the sum of all the digits must be divisible by $9.$ The only way to make this happen is with 9 $4$'s. However, we also need one $9.$

The smallest integer that meets all these conditions is $4444444944$. The last four digits are $\boxed{\mathrm{(C) \ } 4944}$

See Also

2007 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AMC 10 Problems and Solutions

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