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

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Starting with some experimentation, substituting <math>p=5</math> results in <math>24</math>, substituting <math>p=7</math> results in <math>48</math>, and substituting <math>p=11</math> results in <math>120</math>.  For these primes, the resulting numbers are multiples of <math>24</math>.
 
Starting with some experimentation, substituting <math>p=5</math> results in <math>24</math>, substituting <math>p=7</math> results in <math>48</math>, and substituting <math>p=11</math> results in <math>120</math>.  For these primes, the resulting numbers are multiples of <math>24</math>.
  
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To show that all primes <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 1,5 \mod 6</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>.
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To show that all primes we devise the following proof:
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<math>Proof:</math>
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<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 0,1,2,3 \mod 4</math>, <math>p^2 - 1</math> is a multiple of <math>8</math>.  Also, since <math>p^2 \equiv 0,1,2 \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==
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{{AHSME 35p box|year=1973|num-b=17|num-a=19}}
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{{AHSME 30p box|year=1973|num-b=17|num-a=19}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Latest revision as of 13:46, 14 January 2024

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 we devise the following proof:

$Proof:$ $p \ge 5$ 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 $p^2 \equiv 0,1,2,3 \mod 4$, $p^2 - 1$ is a multiple of $8$. Also, since $p^2 \equiv 0,1,2 \mod 3$, $p^2 - 1$ is a multiple of $3$. Thus, $p^2 -1$ is a multiple of $24$, so the answer is $\boxed{\textbf{(C)}}$.

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
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