Difference between revisions of "2005 iTest Problems/Problem 27"
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Find the sum of all non-zero digits that can repeat at the end of a perfect square. (For example, if <math>811</math> were a perfect square, <math>1</math> would be one of these non-zero digits.) | Find the sum of all non-zero digits that can repeat at the end of a perfect square. (For example, if <math>811</math> were a perfect square, <math>1</math> would be one of these non-zero digits.) | ||
==Solution 1== | ==Solution 1== | ||
| − | We seek nonzero digits such that some square ends with the digit repeated. Squares modulo 4 are only 0 or 1, so 11, 22, 55, 66, and 99 are ruled out immediately ( | + | We seek nonzero digits such that some square ends with the digit repeated. Squares modulo 4 are only 0 or 1, so 11, 22, 55, 66, and 99 are ruled out immediately (<math>3,2,3,2,3\mod 4</math>)). Out of the remaining numbers, since a square can have a last digit belonging to the set {<math>0,1,4,5,6,9</math>}, 33, 77, and 88 are ruled out as well. This leaves only 44. Our answer therefore is <math>\boxed{4}</math>. |
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==See Also== | ==See Also== | ||
{{iTest box|year=2005|num-b=26|num-a=28}} | {{iTest box|year=2005|num-b=26|num-a=28}} | ||
[[Category: Intermediate Number Theory Problems]] | [[Category: Intermediate Number Theory Problems]] | ||
Latest revision as of 20:22, 13 October 2025
Problem
Find the sum of all non-zero digits that can repeat at the end of a perfect square. (For example, if 
were a perfect square, 
would be one of these non-zero digits.)
Solution 1
We seek nonzero digits such that some square ends with the digit repeated. Squares modulo 4 are only 0 or 1, so 11, 22, 55, 66, and 99 are ruled out immediately (
)). Out of the remaining numbers, since a square can have a last digit belonging to the set {
}, 33, 77, and 88 are ruled out as well. This leaves only 44. Our answer therefore is 
.
See Also
| 2005 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 26 |
Followed by: Problem 28 | |
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