Difference between revisions of "1983 AIME Problems/Problem 6"

Revision as of 17:51, 16 June 2014

Problem

Let $a_n$ equal $6^{n}+8^{n}$ . Determine the remainder upon dividing $a_ {83}$ by $49$ .

Solution

Solution 1

First, we try to find a relationship between the numbers we're provided with and $49$ . We realize that $49=7^2$ and both $6$ and $8$ are greater or less than $7$ by $1$ .

Expressing the numbers in terms of $7$ , we get $(7-1)^{83}+(7+1)^{83}$ .

Applying the Binomial Theorem, half of our terms cancel out and we are left with $2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)$ . We realize that all of these terms are divisible by $49$ except the final term.

After some quick division, our answer is $\boxed{035}$ .

Solution 2

Since $\phi(49) = 42$ (the Euler's totient function), by Euler's Totient Theorem, $a^{42} \equiv 1 \pmod{49}$ where $\text{gcd}(a,49) = 1$ . Thus $6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1}$ $\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48}$ $\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}$ .

Alternatively, we could have noted that $a^b\equiv a^{b\pmod{\phi{(n)}}}\pmod n$ . This way, we have $6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\pmod {49}$ , and can finish the same way.

@@ Line 18: / Line 18: @@
 \equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>.
-*Alternatively, we could have noted that <math>a^b\equiv a^{b\pmod{\phi{n}}}\pmod n</math>. This way, we have <math>6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\pmod {49}</math>, and can finish the same way.
+*Alternatively, we could have noted that <math>a^b\equiv a^{b\pmod{\phi{(n)}}}\pmod n</math>. This way, we have <math>6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\pmod {49}</math>, and can finish the same way.
 == 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
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