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 13:52, 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 |