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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

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

Solution

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

  • $d<100$, so the maximum $d$ is $99$.
  • $c<4d\le396$, so the maximum $c$ is $395$.
  • $b<3c\le1185$, so the maximum $b$ is $1184$.
  • $a<2b\le2368$, so the maximum $a$ is $\boxed{2367}$.

Note that the statement $a<2b<6c<24d<2400$ is true, but does not specify the distances between each pair of values. These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

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