Difference between revisions of "2018 AMC 10B Problems/Problem 11"
Franchester (talk | contribs) |
Franchester (talk | contribs) (→Solution 2 (Bad Solution)) |
||
| Line 14: | Line 14: | ||
5^2+46=71 \qquad | 5^2+46=71 \qquad | ||
19^2+96=457</math>. | 19^2+96=457</math>. | ||
| − | + | Clearly only <math>\boxed{(\textbf{C})}</math> is our only option left. | |
(franchester) | (franchester) | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2018|ab=B|num-b=10|num-a=12}} | {{AMC10 box|year=2018|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:17, 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.
(franchester)
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. 
