# 1983 AIME Problems/Problem 6

## Problem

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

## Solution

### Solution 1

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

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

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 $\phi(49) = 42$ (see Euler's totient function), Euler's Totient Theorem tells us that $a^{42} \equiv 1 \pmod{49}$ 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.

### Solution 3 (cheap and quick)

As the value of $a$ is obviously $6^{83}+8^{83}$ we look for a pattern with others. With a bit of digging, we discover that $6^n+6^m$ where $m$ and $n$ are odd is equal to $\boxed{35}\:\text{mod}\:49$

-dragoon

### Solution 3 $6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})$

Becuase $7|(6+8)$, we only consider $6^{82}-6^{81}8+\ldots-8^{81}6+8^{82} \pmod{7}$ $6^{82}-6^{81}8+\ldots-8^{81}6+8^{82} \equiv (-1)^{82} - (-1)^{81}+ \ldots - (-1)^1 + 1 = 83 \equiv 6 \pmod{7}$ $6^{83} + 8^{83} \equiv 14 \cdot 6 \equiv \boxed{035} \pmod{49}$

### Solution 4 last resort (bash)

Repeat the steps of taking modulo $49$ after reducing the exponents over and over again until you get a residue of $49,$ namely $35.$ This bashing takes a lot of time but it isn’t too bad. ~peelybonehead

~ pi_is_3.14