Difference between revisions of "1973 AHSME Problems/Problem 18"
(→Solution) |
(→Solution) |
||
Line 16: | Line 16: | ||
<math>Proof:</math> | <math>Proof:</math> | ||
− | <math>p \ge 5</math> result in <math>p^2 -1</math> being a multiple of <math>24</math>, we can use modular arithmetic. Note that <math>p^2 - 1 = (p+1)(p-1)</math>. Since <math>p \equiv 1,3 \mod 4</math>, <math>p^2 - 1</math> is a multiple of <math>8</math>. Also, since <math>p \equiv 1,2 \mod 3</math>, <math>p^2 - 1</math> is a multiple of <math>3</math>. Thus, <math>p^2 -1</math> is a multiple of <math>24</math>, so the answer is <math>\boxed{\textbf{(C)}}</math>. | + | <math>p \ge 5</math> result in <math>p^2 -1</math> being a multiple of <math>24</math>, we can use modular arithmetic. Note that <math>p^2 - 1 = (p+1)(p-1)</math>. Since <math>p^2 \equiv 1,3 \mod 4</math>, <math>p^2 - 1</math> is a multiple of <math>8</math>. Also, since <math>p^2 \equiv 1,2 \mod 3</math>, <math>p^2 - 1</math> is a multiple of <math>3</math>. Thus, <math>p^2 -1</math> is a multiple of <math>24</math>, so the answer is <math>\boxed{\textbf{(C)}}</math>. |
==See Also== | ==See Also== |
Revision as of 14:45, 14 January 2024
Problem
If is a prime number, then
divides
without remainder
Solution
Starting with some experimentation, substituting results in
, substituting
results in
, and substituting
results in
. For these primes, the resulting numbers are multiples of
.
To show that all primes we devise the following proof:
result in
being a multiple of
, we can use modular arithmetic. Note that
. Since
,
is a multiple of
. Also, since
,
is a multiple of
. Thus,
is a multiple of
, so the answer is
.
See Also
1973 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |