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 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 <math>\textbf{(D)}</math> | + | 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>. | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1957|num-b=37|num-a=39}} | ||
+ | {{MAA Notice}} |
Latest revision as of 08:35, 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 | |
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