Difference between revisions of "1952 AHSME Problems/Problem 44"
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Let <math>n = 10a+b</math>. The problem states that <math>10a+b=k(a+b)</math>. We want to find <math>x</math>, where <math>10b+a=x(a+b)</math>. Adding these two equations gives <math>11(a+b) = (k+x)(a+b)</math>. Because <math>a+b \neq 0</math>, we have <math>11 = k + x</math>, or <math>x = \boxed{\textbf{(C) \ } 11-k}</math>. | Let <math>n = 10a+b</math>. The problem states that <math>10a+b=k(a+b)</math>. We want to find <math>x</math>, where <math>10b+a=x(a+b)</math>. Adding these two equations gives <math>11(a+b) = (k+x)(a+b)</math>. Because <math>a+b \neq 0</math>, we have <math>11 = k + x</math>, or <math>x = \boxed{\textbf{(C) \ } 11-k}</math>. | ||
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== See Also == | == See Also == | ||
{{AHSME 50p box|year=1952|num-b=43|num-a=45}} | {{AHSME 50p box|year=1952|num-b=43|num-a=45}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:57, 24 December 2015
Problem
If an integer of two digits is 
times the sum of its digits, the number formed by interchanging the the digits is the sum of the digits multiplied by
Solution
Let 
. The problem states that 
. We want to find 
, where 
. Adding these two equations gives 
. Because 
, we have 
, or 
.
See Also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 43 |
Followed by Problem 45 | |
| 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. 
