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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 06:49, 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 (Problems • Answer Key • Resources)
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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