Difference between revisions of "1957 AHSME Problems/Problem 38"
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| + | ==Problem== | ||
| + | From a two-digit number <math>N</math> we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then: | ||
| + | |||
| + | <math>\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}</math> | ||
==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)} | ||
==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 
we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:
Solution
The number can be written as 
with 
and 
representing the digits. The number with its digits reversed is 
. Since the problem asks for a positive number as the difference of these two numbers, than 
. Writing this out, we get 
. Therefore, the difference must be a multiple of 
, and the only perfect cube with less than 
digits and is multiple of is 
. Also, that means 
, and there are 
possibilities of that, so our answer is
.
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. 
