Difference between revisions of "2018 AMC 10B Problems/Problem 11"
Awesomemaths (talk | contribs) (→Solution 1) |
|||
| Line 6: | Line 6: | ||
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: Isn't it <math>E</math>? <math>11^2 +26 = 147</math> which is divisible by <math>3</math> | ||
==See Also== | ==See Also== | ||
Revision as of 16:26, 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: Isn't it 
? 
which is divisible by
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
