Difference between revisions of "1973 AHSME Problems/Problem 18"

Revision as of 13:00, 20 February 2020

Problem

If $p \geq 5$ is a prime number, then $24$ divides $p^2 - 1$ without remainder

$\textbf{(A)}\ \text{never} \qquad \textbf{(B)}\ \text{sometimes only} \qquad \textbf{(C)}\ \text{always} \qquad$ $\textbf{(D)}\ \text{only if } p =5 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Starting with some experimentation, substituting $p=5$ results in $24$ , substituting $p=7$ results in $48$ , and substituting $p=11$ results in $120$ . For these primes, the resulting numbers are multiples of $24$ .

To show that all primes $p \ge 5$ result in $p^2 -1$ being a multiple of $24$ , we can use modular arithmetic. Note that $p^2 - 1 = (p+1)(p-1)$ . Since $p \equiv 1,3 \mod 4$ , $p^2 - 1$ is a multiple of $8$ . Also, since $p \equiv 1,5 \mod 6$ , $p^2 - 1$ is a multiple of $3$ . Thus, $p^2 -1$ is a multiple of $24$ , so the answer is $\boxed{\textbf{(C)}}$ .

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

@@ Line 16: / Line 16: @@
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=17|num-a=19}}
+{{AHSME 30p box|year=1973|num-b=17|num-a=19}}
 [[Category:Introductory Number Theory Problems]]

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Difference between revisions of "1973 AHSME Problems/Problem 18"

Revision as of 13:00, 20 February 2020

Problem

Solution

See Also