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 "1957 AHSME Problems/Problem 38"

(→‎Solution)
m (typo fix/formatting change)
 
Line 5: Line 5:
 
==Solution==
 
==Solution==
 
The number <math>N</math> can be written as <math>10a+b</math> with <math>a</math> and <math>b</math> representing the digits. The number <math>N</math> with its digits reversed is <math>10b+a</math>. Since the problem asks for a positive number as the difference of these two numbers, than <math>a>b</math>. Writing this out, we get <math>10a+b-(10b+a)=9a-9b=9(a-b)</math>. Therefore, the difference must be a multiple of <math>9</math>, and the only perfect cube with less than <math>3</math> digits and is multiple of <math>9</math> is <math>3^3=27</math>. Also, that means <math>a-b=3</math>, and there are <math>7</math> possibilities of that, so our answer is  
 
The number <math>N</math> can be written as <math>10a+b</math> with <math>a</math> and <math>b</math> representing the digits. The number <math>N</math> with its digits reversed is <math>10b+a</math>. Since the problem asks for a positive number as the difference of these two numbers, than <math>a>b</math>. Writing this out, we get <math>10a+b-(10b+a)=9a-9b=9(a-b)</math>. Therefore, the difference must be a multiple of <math>9</math>, and the only perfect cube with less than <math>3</math> digits and is multiple of <math>9</math> is <math>3^3=27</math>. Also, that means <math>a-b=3</math>, and there are <math>7</math> possibilities of that, so our answer is  
−
 
+
<math>\boxed{\textbf{(D) } \text{There are exactly } 7 \text{ values for } N}</math>.
−
<math>\boxed{\textbf{(D)}}</math> There are exactly <math>7</math> values of <math>N</math>
 
  
 
==See Also==
 
==See Also==
  
−
{{AHSME box|year=1957|num-b=37|num-a=39}}
+
{{AHSME 50p box|year=1957|num-b=37|num-a=39}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 07:35, 27 July 2024

Problem

From a two-digit number $N$ we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:

$\textbf{(A)}\ {N}\text{ cannot end in 5}\qquad\\ \textbf{(B)}\ {N}\text{ can end in any digit other than 5}\qquad \\ \textbf{(C)}\ {N}\text{ does not exist}\qquad\\ \textbf{(D)}\ \text{there are exactly 7 values for }{N}\qquad\\ \textbf{(E)}\ \text{there are exactly 10 values for }{N}$

Solution

The number $N$ can be written as $10a+b$ with $a$ and $b$ representing the digits. The number $N$ with its digits reversed is $10b+a$. Since the problem asks for a positive number as the difference of these two numbers, than $a>b$. Writing this out, we get $10a+b-(10b+a)=9a-9b=9(a-b)$. Therefore, the difference must be a multiple of $9$, and the only perfect cube with less than $3$ digits and is multiple of $9$ is $3^3=27$. Also, that means $a-b=3$, and there are $7$ possibilities of that, so our answer is $\boxed{\textbf{(D) } \text{There are exactly } 7 \text{ values for } N}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 37 		Followed by
Problem 39
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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=1957_AHSME_Problems/Problem_38&oldid=223750"