Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | <math>\{2,4 | + | Factoring, we get <math>n^3-12n^2+40n-29 = (n-1)(n^2-11n+29)</math>. Thus, we must have that either <math>n-1</math> or <math>n^2-11n+29</math> equal to <math>1</math>. If we have <math>n-1</math> equal to 1, we have <math>n=2</math>. Plugging back in the polynomial, we get <math>11</math>, which is a prime, so <math>2</math> works. If <math>n^2-11n+29</math> is equal to one, we have <math>n^2-11n+28=0</math>, so <math>n=4</math> or <math>n=7</math>. Plugging both back in the polynomial, we get <math>3</math> and <math>6</math>, respectively. <math>3</math> is a prime, but <math>6</math> is not, so <math>4</math> works. Thus, the answer is <math>\boxed{2,4}</math> |
==See Also== | ==See Also== | ||
Latest revision as of 12:52, 9 August 2023
Problem
Find all positive integers 
such that 
is a prime number. For each of your values of compute
this cubic polynomial showing that it is, in fact, a prime.
Solution
Factoring, we get 
. Thus, we must have that either 
or 
equal to 
. If we have equal to 1, we have 
. Plugging back in the polynomial, we get 
, which is a prime, so 
works. If is equal to one, we have 
, so 
or 
. Plugging both back in the polynomial, we get 
and 
, respectively. is a prime, but is not, so 
works. Thus, the answer is 
See Also
| 2006 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
