1989 AHSME Problems/Problem 11

Problem

Let $a$ , $b$ , $c$ , and $d$ be positive integers with $a < 2b$ , $b < 3c$ , and $c<4d$ . If $d<100$ , the largest possible value for $a$ is

$\textrm{(A)}\ 2367\qquad\textrm{(B)}\ 2375\qquad\textrm{(C)}\ 2391\qquad\textrm{(D)}\ 2399\qquad\textrm{(E)}\ 2400$

Each of these integers is bounded above by the next one.

Note that the statement $a<2b<6c<24d<2400$ is true, but does not specify the distances between each pair of values.

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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