Difference between revisions of "1989 AHSME Problems/Problem 11"

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Note that the statement <math>a<2b<6c<24d<2400</math> is true, but does not specify the distances between each pair of values.
 
Note that the statement <math>a<2b<6c<24d<2400</math> is true, but does not specify the distances between each pair of values.
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== See also ==
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{{AHSME box|year=1989|num-b=10|num-a=12}} 
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[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 07:49, 22 October 2014

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.

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 26 27 28 29 30
All AHSME Problems and Solutions

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