1973 AMC 12 Problem 18

Attempt Problem 18 of the 1973 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1973 AMC 12 solutions, or check the answer key.

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18.

If p5p\ge5 is a prime number, then 2424 divides p21p^2-1 without remainder

never

sometimes only

always

only if p=5p=5

none of these

Answer: C
Concepts:divisibilityprimemodular arithmetic
Difficulty rating: 1610
Small Hint:

Factor p21=(p1)(p+1)p^2-1=(p-1)(p+1)

Big Hint:

Among the three consecutive integers p1,p-1, p,p, p+1,p+1, locate factors of 33 and 88

Solution:

A prime p5p\ge5 is odd, so p1p-1 and p+1p+1 are consecutive even integers. One is divisible by 4,4, making their product divisible by 8.8. Among the three consecutive integers p1,p-1, p,p, p+1,p+1, one is divisible by 3.3. It cannot be p,p, since p5p\ge5 is prime, so 33 also divides (p1)(p+1).(p-1)(p+1). Because 33 and 88 are relatively prime, (p1)(p+1)=p21 (p-1)(p+1)=p^2-1 is always divisible by 24.24.

Therefore, the correct answer is C.

← Problem 17#17
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