Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "== Problem == Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be positive integers with <math> a < 2b</math>, <math>b < 3c </math>, and <math>c<4d</math>....")
 
Line 15: Line 15:
  
 
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.
 +
{{MAA Notice}}

Revision as of 12:48, 5 July 2013

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. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_11&oldid=56378"