1997 AHSME Problems/Problem 13: Difference between revisions
Talkinaway (talk | contribs) No edit summary |
No edit summary |
||
| Line 19: | Line 19: | ||
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=12|num-a=14}} | {{AHSME box|year=1997|num-b=12|num-a=14}} | ||
{{MAA Notice}} | |||
Latest revision as of 13:12, 5 July 2013
Problem
How many two-digit positive integers 
have the property that the sum of and the number obtained by reversing the order of the digits of is a perfect square?
Solution
Let 
, where 
is the tens digit and 
is the units digit.
The condition of the problem is that 
is a perfect square.
Simplifying and factoring, we want 
to be a perfect square.
Thus, 
must at least be a multiple of 
, and since and are digits, the only multiple of that works is itself.
Thus, 
is the first solution, and 
is the last solution. There are 
solutions in total, leading to answer 
.
See also
| 1997 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
