Difference between revisions of "1957 AHSME Problems/Problem 38"
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==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 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> | ||
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| + | ==See Also== | ||
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| + | {{AHSME box|year=1957|num-b=37|num-a=39}} | ||
| + | {{MAA Notice}} | ||
Revision as of 20:02, 18 June 2019
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 
See Also
| 1957 AHSME (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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
