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

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== Problem ==
 
== Problem ==
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Let <math>a_n</math> equal <math>6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>.
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Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>.
  
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__TOC__
 
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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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>.  
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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>.  
  
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Expressing the numbers in terms of <math>7</math>, we get <math>(7-1)^{83}+(7+1)^{83}</math>.
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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.
 
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.
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=== Solution 2 ===
 
=== Solution 2 ===
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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>  
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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>
 
<math>\equiv 6^{-1} + 8^{-1} \equiv \frac{8+6}{48} </math> <math>
 
\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>.
 
\equiv \frac{14}{-1}\equiv \boxed{035} \pmod{49}</math>.
  
 
*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.
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=== Solution 3 (cheap and quick) ===
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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+6^m</math> where <math>m</math> and <math>n</math> are odd is equal to <math>\boxed{35}\:\text{mod}\:49</math>
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-dragoon
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=== Solution 3===
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<math>6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})</math>
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Becuase <math>7|(6+8)</math>, we only consider  <math>6^{82}-6^{81}8+\ldots-8^{81}6+8^{82}  \pmod{7}</math>
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<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>
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<math>6^{83} + 8^{83} \equiv 14 \cdot 6 \equiv \boxed{035} \pmod{49}</math>
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=== Solution 4 last resort (bash) ===
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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
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== Video Solution by OmegaLearn ==
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https://youtu.be/-H4n-QplQew?t=792
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~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==

Latest revision as of 19:20, 14 January 2023

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

Video Solution by OmegaLearn

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

~ pi_is_3.14

See Also

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