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

(Cleaned up the solutions)
Line 19: Line 19:
  
 
*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.
 +
 +
== Solution 3==
 +
<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}) \text{mod} 7</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 12:52, 9 March 2021

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

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

Becuase $7|(6+8)$, we only consider $(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82}) \text{mod} 7$

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