1973 AHSME Problems/Problem 18
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 | |
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