Difference between revisions of "1987 AJHSME Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | "If a whole number <math>n</math> is not | + | "If a whole number <math>n</math> is not prime, then the whole number <math>n-2</math> is not prime." A value of <math>n</math> which shows this statement to be false is |
− | <math>\ | + | <math>\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 23</math> |
==Solution== | ==Solution== | ||
Line 11: | Line 11: | ||
Now we just check the statement for <math>n=9,12,16</math>. If <math>n=12</math> or <math>n=16</math>, then <math>n-2</math> is <math>10</math> or <math>14</math>, which aren't prime. However, <math>n=9</math> makes <math>n-2=7</math>, which is prime, so <math>n=9</math> proves the statement false. | Now we just check the statement for <math>n=9,12,16</math>. If <math>n=12</math> or <math>n=16</math>, then <math>n-2</math> is <math>10</math> or <math>14</math>, which aren't prime. However, <math>n=9</math> makes <math>n-2=7</math>, which is prime, so <math>n=9</math> proves the statement false. | ||
− | <math>\boxed{\ | + | Therefore, the answer is <math>\boxed{\textbf{A}}</math>, 9. |
==See Also== | ==See Also== |
Latest revision as of 16:44, 24 May 2023
Problem
"If a whole number is not prime, then the whole number
is not prime." A value of
which shows this statement to be false is
Solution
To show this statement to be false, we need a non-prime value of such that
is prime. Since
and
are prime, they won't prove anything relating to the truth of the statement.
Now we just check the statement for . If
or
, then
is
or
, which aren't prime. However,
makes
, which is prime, so
proves the statement false.
Therefore, the answer is , 9.
See Also
1987 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.