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

(Solution 1)
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\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>.
 
\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>.
  
== See also ==
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== See Also ==
 
{{AIME box|year=1983|num-b=5|num-a=7}}
 
{{AIME box|year=1983|num-b=5|num-a=7}}
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 06:01, 16 April 2012

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 $8$ 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 $6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1}$ $\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48}$ $\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}$.

See Also

1983 AIME (ProblemsAnswer KeyResources)
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