1983 AIME Problems/Problem 6
Revision as of 01:47, 21 April 2021 by Otms mathclub (talk | contribs) (83 is clearly not congruent to 6 mod 49.)
Problem
Let . Determine the remainder upon dividing
by
.
Solution
Solution 1
Firstly, we try to find a relationship between the numbers we're provided with and . We notice that
, and both
and
are greater or less than
by
.
Thus, expressing the numbers in terms of , we get
.
Applying the Binomial Theorem, half of our terms cancel out and we are left with . We realize that all of these terms are divisible by
except the final term.
After some quick division, our answer is .
Solution 2
Since (see Euler's totient function), Euler's Totient Theorem tells us that
where
. Thus
.
- Alternatively, we could have noted that
. This way, we have
, and can finish the same way.
Solution 3
Becuase , we only consider
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |