1987 AJHSME Problems/Problem 10

Problem

$4(299)+3(299)+2(299)+298=$

$\text{(A)}\ 2889 \qquad \text{(B)}\ 2989 \qquad \text{(C)}\ 2991 \qquad \text{(D)}\ 2999 \qquad \text{(E)}\ 3009$

Solution

We can make use of the distributive property as follows: $\begin{align*} 4(299)+3(299)+2(299)+298 &= 4(299)+3(299)+2(299)+1(299)-1 \\ &= (4+3+2+1)(299)-1 \\ &= 10(299)-1 \\ &= 2989 \\ \end{align*}$

$\boxed{\text{B}}$

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AJHSME/AMC 8 Problems and Solutions

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1987 AJHSME Problems/Problem 10

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