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

Latest revision as of 08:08, 31 August 2025

Problem

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

Solutions

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

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

Because $7|(6+8)$ , 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

Video Solution by OmegaLearn

https://youtu.be/-H4n-QplQew?t=792

~ pi_is_3.14

@@ Line 21: / Line 21: @@
 *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.
-=== Solution 3 (cheap and quick) ===
+=== Solution 3 ===
-As the value of <math>a</math> is obviously <math>6^{83}+8^{83}</math> we look for a pattern with others. With a bit of digging, we discover that <math>6^n+8^m</math> where <math>m</math> and <math>n</math> are odd is equal to <math>\boxed{35}\:\text{mod}\:49</math>
--dragoon
--minor edits by flyingkinder123
-Note:
-This isn't true??
-Take <math>n=m=1,3,5,7</math>
-XD
-=== Solution 4===
 <math>6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})</math>
-Becuase <math>7|(6+8)</math>, we only consider  <math>6^{82}-6^{81}8+\ldots-8^{81}6+8^{82}  \pmod{7}</math>
+Because <math>7|(6+8)</math>, we only consider  <math>6^{82}-6^{81}8+\ldots-8^{81}6+8^{82}  \pmod{7}</math>
 <math>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}</math>
@@ Line 45: / Line 32: @@
-=== Solution 5 last resort (bash) ===
+=== Solution 4 last resort (bash) ===
 Repeat the steps of taking modulo <math>49</math> after reducing the exponents over and over again until you get a residue of <math>49,</math> namely <math>35.</math> This bashing takes a lot of time but it isn’t too bad. ~peelybonehead

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