Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 4"
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Determine all positive integers <math>n</math> such that <math>n^2+3</math> divides evenly (without remainder) into <math>n^4-3n^2+10</math> ? | Determine all positive integers <math>n</math> such that <math>n^2+3</math> divides evenly (without remainder) into <math>n^4-3n^2+10</math> ? | ||
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==Solution== | ==Solution== | ||
− | {{ | + | By polynomial division, <math>\frac{n^4 - 3n^2 + 10}{n^2 + 3} = n^2 - 6 + \frac{28}{n^2 + 3}</math>. By default, <math>n^2 - 6 \in \mathbb{N}</math> when <math>n \in \mathbb{N}</math>, so we must find all positive integral values of <math>n</math> for which <math>\frac{28}{n^2 + 3}</math> is also an integer. Listing out the factors of <math>28</math>, we see that <math>n^2 + 3 \in \{1, 2, 4, 7, 14, 28\}</math>. Looking for positive integral values that satisfy these conditions yields <math>\boxed{n \in \{1, 2, 5\}}</math>, as desired. <math>\blacksquare</math> |
==See Also== | ==See Also== |
Latest revision as of 16:07, 18 December 2018
Problem
Determine all positive integers such that divides evenly (without remainder) into ?
Solution
By polynomial division, . By default, when , so we must find all positive integral values of for which is also an integer. Listing out the factors of , we see that . Looking for positive integral values that satisfy these conditions yields , as desired.
See Also
2006 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |