1976 AHSME Problems/Problem 15
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Problem 15
If is the remainder when each of the numbers , and is divided by , where is an integer greater than , then equals
Solution
We are given these congruences:
Let's make a new congruence by subtracting (i) from (ii), which results in
Subtract (ii) from (iii) to get
Now we know that and are both multiples of . Their prime factorizations are and , so their common factor is , which means .
Plug back into any of the original congruences to get . Then, . ~jiang147369
See Also
1976 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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