Difference between revisions of "2018 AMC 10B Problems/Problem 11"
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Because squares of a non-multiple of 3 is always <math>1\mod 3</math>, the only expression is always a multiple of <math>3</math> is <math>\boxed{\textbf{(C) } p^2+26} </math>. This is excluding when <math>p=0\mod3</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite. | Because squares of a non-multiple of 3 is always <math>1\mod 3</math>, the only expression is always a multiple of <math>3</math> is <math>\boxed{\textbf{(C) } p^2+26} </math>. This is excluding when <math>p=0\mod3</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite. | ||
+ | |||
+ | ==Solution 2 (Bad Solution)== | ||
+ | |||
+ | We proceed with guess and check: | ||
+ | <math>5^2+16=41</math> | ||
+ | <math>7^2+24=73</math> | ||
+ | <math>11^2+26=147</math> | ||
+ | <math>5^2+46=71</math> | ||
+ | Clearly only <math>\textbf{E}</math> is our only option left. | ||
==See Also== | ==See Also== |
Revision as of 17:12, 16 February 2018
Which of the following expressions is never a prime number when is a prime number?
Solution 1
Because squares of a non-multiple of 3 is always , the only expression is always a multiple of is . This is excluding when , which only occurs when , then which is still composite.
Solution 2 (Bad Solution)
We proceed with guess and check: Clearly only is our only option left.
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 | ||
All AMC 10 Problems and Solutions |
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