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:
![$1059 \equiv r \pmod{d} \qquad$](http://latex.artofproblemsolving.com/1/7/0/170f066e38c088653b2597e44e9980b3d4b03896.png)
![$1417 \equiv r \pmod{d} \qquad$](http://latex.artofproblemsolving.com/2/c/b/2cb69c85eef250e85dec2aeef4578e59c735f228.png)
![$2312 \equiv r \pmod{d} \qquad$](http://latex.artofproblemsolving.com/3/4/b/34be5ab4060e3f47479db32d51cc629594eb4315.png)
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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