1983 AIME Problems/Problem 6: Difference between revisions

Revision as of 18:24, 15 February 2019

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.

@@ Line 5: / Line 5: @@
 == Solution ==
 === Solution 1 ===
-First, we try to find a relationship between the numbers we're provided with and <math>49</math>. We realize that <math>49=7^2</math> and both <math>6</math> and <math>8</math> are greater or less than <math>7</math> by <math>1</math>.
+Firstly, we try to find a relationship between the numbers we're provided with and <math>49</math>. We notice that <math>49=7^2</math>, and both <math>6</math> and <math>8</math> are greater or less than <math>7</math> by <math>1</math>.
-Expressing the numbers in terms of <math>7</math>, we get <math>(7-1)^{83}+(7+1)^{83}</math>.
+Thus, expressing the numbers in terms of <math>7</math>, we get <math>a_{83} = (7-1)^{83}+(7+1)^{83}</math>.
 Applying the [[Binomial Theorem]], half of our terms cancel out and we are left with <math>2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)</math>. We realize that all of these terms are divisible by <math>49</math> except the final term.
@@ Line 14: / Line 14: @@
 === Solution 2 ===
-Since <math>\phi(49) = 42</math> (the [[Euler's totient function]]), by [[Euler's Totient Theorem]], <math>a^{42} \equiv 1 \pmod{49}</math> where <math>\text{gcd}(a,49) = 1</math>. Thus <math>6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1} </math>
+Since <math>\phi(49) = 42</math> (see [[Euler's totient function]]), [[Euler's Totient Theorem]] tells us that <math>a^{42} \equiv 1 \pmod{49}</math> where <math>\text{gcd}(a,49) = 1</math>. Thus <math>6^{83} + 8^{83} \equiv 6^{2(42)-1}+8^{2(42)-1} </math>
 <math>\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48} </math> <math>
 \equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>.

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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