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 "1998 AHSME Problems/Problem 6"
 
(One intermediate revision by one other user not shown)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
{{solution}}
+
 
 +
If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to <math>\sqrt{1998}</math> as possible.
 +
 
 +
Since <math>45^2 = 2025</math>, the factors should be as close to <math>44</math> or <math>45</math> as possible.
 +
 
 +
Breaking down <math>1998</math> into its prime factors gives <math>1998 = 2\cdot 3^3 \cdot 37</math>. 
 +
 
 +
<math>37</math> is relatively close to <math>44</math>, and no numbers between <math>38</math> and <math>44</math> are factors of <math>1998</math>.  Thus, the two factors are <math>37</math> and <math>2\cdot 3^3 = 54</math>, and the difference is <math>54 - 37 = 17</math>, and the answer is <math>\boxed{C}</math>
 +
 
 
==See Also==
 
==See Also==
  
 
{{AHSME box|year=1998|num-b=5|num-a=7}}
 
{{AHSME box|year=1998|num-b=5|num-a=7}}
 +
{{MAA Notice}}

Latest revision as of 13:28, 5 July 2013

Problem

If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is

$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }17 \qquad \mathrm{(D) \ }47 \qquad \mathrm{(E) \ } 93$

Solution

If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to $\sqrt{1998}$ as possible.

Since $45^2 = 2025$, the factors should be as close to $44$ or $45$ as possible.

Breaking down $1998$ into its prime factors gives $1998 = 2\cdot 3^3 \cdot 37$.

$37$ is relatively close to $44$, and no numbers between $38$ and $44$ are factors of $1998$. Thus, the two factors are $37$ and $2\cdot 3^3 = 54$, and the difference is $54 - 37 = 17$, and the answer is $\boxed{C}$

See Also

1998 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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

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=1998_AHSME_Problems/Problem_6&oldid=56536"