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Difference between revisions of "2021 AIME II Problems/Problem 9"

m (→‎Solution 1: Minor edits on uppercases. Also, I reformatted the equation block so that it appears much nicer.)
(Prioritize the comprehensive solution. Also, made the solution more result-driven by separating the solutions and the important results.)
Line 3: Line 3:
  
 
==Solution 1==
 
==Solution 1==
 +
This solution refers to the <b>Remarks</b> section.
 +
 +
By the Euclidean Algorithm, we have <cmath>\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.</cmath>
 +
We are given that <math>\gcd\left(2^m+1,2^n-1\right)>1.</math> Multiplying both sides by <math>\gcd\left(2^m-1,2^n-1\right)</math> gives
 +
<cmath>\begin{align*}
 +
\gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\
 +
\gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \hspace{12mm} &&\text{by }\textbf{Claim 2} \\
 +
\gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\
 +
2^{\gcd(2m,n)}-1&>2^{\gcd(m,n)}-1 &&\text{by }\textbf{Claim 1} \\
 +
\gcd(2m,n)&>\gcd(m,n),
 +
\end{align*}</cmath>
 +
which implies that <math>n</math> must have more factors of <math>2</math> than <math>m</math> does.
 +
 +
We construct the following table for the first <math>30</math> positive integers:
 +
<cmath>\begin{array}{c|c|c}
 +
&& \\ [-2.5ex]
 +
\boldsymbol{\#}\textbf{ of Factors of }\boldsymbol{2} & \textbf{Numbers} & \textbf{Count} \\
 +
\hline
 +
&& \\ [-2.25ex]
 +
0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\
 +
&& \\ [-2.25ex]
 +
1 & 2,6,10,14,18,22,26,30 & 8 \\
 +
&& \\ [-2.25ex]
 +
2 & 4,12,20,28 & 4 \\ 
 +
&& \\ [-2.25ex]
 +
3 & 8,24 & 2 \\
 +
&& \\ [-2.25ex]
 +
4 & 16 & 1 \\
 +
\end{array}</cmath>
 +
To count the ordered pairs <math>(m,n),</math> we perform casework on the number of factors of <math>2</math> that <math>m</math> has:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>If <math>m</math> has <math>0</math> factors of <math>2,</math> then <math>m</math> has <math>15</math> options and <math>n</math> has <math>8+4+2+1=15</math> options. So, this case has <math>15\cdot15=225</math> ordered pairs.</li><p>
 +
  <li>If <math>m</math> has <math>1</math> factor of <math>2,</math> then <math>m</math> has <math>8</math> options and <math>n</math> has <math>4+2+1=7</math> options. So, this case has <math>8\cdot7=56</math> ordered pairs.</li><p>
 +
  <li>If <math>m</math> has <math>2</math> factors of <math>2,</math> then <math>m</math> has <math>4</math> options and <math>n</math> has <math>2+1=3</math> options. So, this case has <math>4\cdot3=12</math> ordered pairs.</li><p>
 +
  <li>If <math>m</math> has <math>3</math> factors of <math>2,</math> then <math>m</math> has <math>2</math> options and <math>n</math> has <math>1</math> option. So, this case has <math>2\cdot1=2</math> ordered pairs.</li>
 +
</ol>
 +
Together, the answer is <math>225+56+12+2=\boxed{295}.</math>
 +
 +
~MRENTHUSIASM
 +
 +
==Solution 2==
 
We make use of the (Olympiad Number Theory) lemma that <math>\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1</math>.
 
We make use of the (Olympiad Number Theory) lemma that <math>\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1</math>.
  
Line 16: Line 57:
 
~Lcz
 
~Lcz
  
−
==Solution 2 (Comprehensive and Generalized)==
+
==Remarks==
−
===Claim (Solution 1's Lemma)===
+
===Claim 1 (Olympiad Number Theory Lemma)===
 
If <math>u,a,</math> and <math>b</math> are positive integers such that <math>u\geq2,</math> then <math>\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.</math>
 
If <math>u,a,</math> and <math>b</math> are positive integers such that <math>u\geq2,</math> then <math>\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.</math>
  
Line 24: Line 65:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
===Proof 1 (Euclidean Algorithm)===
+
===Claim 1 Proof 1 (Euclidean Algorithm)===
 
If <math>a=b,</math> then <math>\gcd(a,b)=a=b,</math> from which the claim is clearly true.
 
If <math>a=b,</math> then <math>\gcd(a,b)=a=b,</math> from which the claim is clearly true.
  
Line 40: Line 81:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
===Proof 2 (Bézout's Identity)===
+
===Claim 1 Proof 2 (Bézout's Identity)===
 
Let <math>d=\gcd\left(u^a-1,u^b-1\right).</math> It follows that <math>u^a\equiv1\pmod{d}</math> and <math>u^b\equiv1\pmod{d}.</math>
 
Let <math>d=\gcd\left(u^a-1,u^b-1\right).</math> It follows that <math>u^a\equiv1\pmod{d}</math> and <math>u^b\equiv1\pmod{d}.</math>
  
Line 55: Line 96:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
===Solution (Detailed Explanation of Solution 1)===
+
===Claim 2 (GCD Property)===
−
By the Euclidean Algorithm, we have <cmath>\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1. \hspace{41.125mm} (*)</cmath>
 
−
We are given that <math>\gcd\left(2^m+1,2^n-1\right)>1.</math> Multiplying both sides by <math>\gcd\left(2^m-1,2^n-1\right)</math> gives
 
−
<cmath>\begin{align*}
 
−
\gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \hspace{5.25mm} (**) \\
 
−
\underbrace{\gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)}_{\text{by }(*)\text{ and }\textbf{Remark (GCD Property)}}&>\gcd\left(2^m-1,2^n-1\right) \\
 
−
\gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\
 
−
\underbrace{2^{\gcd(2m,n)}-1}_{\text{by }\textbf{Claim}}&>\underbrace{2^{\gcd(m,n)}-1}_{\text{by }\textbf{Claim}} \\
 
−
\gcd(2m,n)&>\gcd(m,n),
 
−
\end{align*}</cmath>
 
−
which implies that <math>n</math> must have more factors of <math>2</math> than <math>m</math> does.
 
−
 
 
−
We construct the following table for the first <math>30</math> positive integers:
 
−
<cmath>\begin{array}{c|c|c}
 
−
&& \\ [-2.5ex]
 
−
\boldsymbol{\#}\textbf{ of Factors of }\boldsymbol{2} & \textbf{Numbers} & \textbf{Count} \\
 
−
\hline
 
−
&& \\ [-2.25ex]
 
−
0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\
 
−
&& \\ [-2.25ex]
 
−
1 & 2,6,10,14,18,22,26,30 & 8 \\
 
−
&& \\ [-2.25ex]
 
−
2 & 4,12,20,28 & 4 \\ 
 
−
&& \\ [-2.25ex]
 
−
3 & 8,24 & 2 \\
 
−
&& \\ [-2.25ex]
 
−
4 & 16 & 1 \\
 
−
\end{array}</cmath>
 
−
To count the ordered pairs <math>(m,n),</math> we perform casework on the number of factors of <math>2</math> that <math>m</math> has:
 
−
<ol style="margin-left: 1.5em;">
 
−
  <li>If <math>m</math> has <math>0</math> factors of <math>2,</math> then <math>m</math> has <math>15</math> options and <math>n</math> has <math>8+4+2+1=15</math> options. So, this case has <math>15\cdot15=225</math> ordered pairs.</li><p>
 
−
  <li>If <math>m</math> has <math>1</math> factor of <math>2,</math> then <math>m</math> has <math>8</math> options and <math>n</math> has <math>4+2+1=7</math> options. So, this case has <math>8\cdot7=56</math> ordered pairs.</li><p>
 
−
  <li>If <math>m</math> has <math>2</math> factors of <math>2,</math> then <math>m</math> has <math>4</math> options and <math>n</math> has <math>2+1=3</math> options. So, this case has <math>4\cdot3=12</math> ordered pairs.</li><p>
 
−
  <li>If <math>m</math> has <math>3</math> factors of <math>2,</math> then <math>m</math> has <math>2</math> options and <math>n</math> has <math>1</math> option. So, this case has <math>2\cdot1=2</math> ordered pairs.</li>
 
−
</ol>
 
−
Together, the answer is <math>225+56+12+2=\boxed{295}.</math>
 
−
 
 
−
~MRENTHUSIASM
 
−
 
 
−
===Remark (GCD Property)===
 
−
In <math>(**)</math> of the solution, we use the following property of the greatest common divisor:
 
−
 
 
 
For positive integers <math>r,s,</math> and <math>t,</math> if <math>\gcd(r,s)=1,</math> then <math>\gcd(r,t)\cdot\gcd(s,t)=\gcd(rs,t).</math>
 
For positive integers <math>r,s,</math> and <math>t,</math> if <math>\gcd(r,s)=1,</math> then <math>\gcd(r,t)\cdot\gcd(s,t)=\gcd(rs,t).</math>
  

Revision as of 09:54, 26 August 2021

Problem

Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$.

Solution 1

This solution refers to the Remarks section.

By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\] We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives \begin{align*} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ \gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \hspace{12mm} &&\text{by }\textbf{Claim 2} \\ \gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ 2^{\gcd(2m,n)}-1&>2^{\gcd(m,n)}-1 &&\text{by }\textbf{Claim 1} \\ \gcd(2m,n)&>\gcd(m,n), \end{align*} which implies that $n$ must have more factors of $2$ than $m$ does.

We construct the following table for the first $30$ positive integers: \[\begin{array}{c|c|c} && \\ [-2.5ex] \boldsymbol{\#}\textbf{ of Factors of }\boldsymbol{2} & \textbf{Numbers} & \textbf{Count} \\ \hline && \\ [-2.25ex] 0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\ && \\ [-2.25ex] 1 & 2,6,10,14,18,22,26,30 & 8 \\ && \\ [-2.25ex] 2 & 4,12,20,28 & 4 \\ && \\ [-2.25ex] 3 & 8,24 & 2 \\ && \\ [-2.25ex] 4 & 16 & 1 \\ \end{array}\] To count the ordered pairs $(m,n),$ we perform casework on the number of factors of $2$ that $m$ has:

  1. If $m$ has $0$ factors of $2,$ then $m$ has $15$ options and $n$ has $8+4+2+1=15$ options. So, this case has $15\cdot15=225$ ordered pairs.

  2. If $m$ has $1$ factor of $2,$ then $m$ has $8$ options and $n$ has $4+2+1=7$ options. So, this case has $8\cdot7=56$ ordered pairs.

  3. If $m$ has $2$ factors of $2,$ then $m$ has $4$ options and $n$ has $2+1=3$ options. So, this case has $4\cdot3=12$ ordered pairs.

  4. If $m$ has $3$ factors of $2,$ then $m$ has $2$ options and $n$ has $1$ option. So, this case has $2\cdot1=2$ ordered pairs.

Together, the answer is $225+56+12+2=\boxed{295}.$

~MRENTHUSIASM

Solution 2

We make use of the (Olympiad Number Theory) lemma that $\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1$.

Noting $\gcd(2^m+1,2^m-1)=\gcd(2^m+1,2)=1$, we have (by difference of squares) \begin{align*} \gcd(2^m+1,2^n-1) \neq 1 &\iff \gcd(2^{2m}-1,2^n-1) \neq \gcd(2^m-1,2^n-1) \\ &\iff 2^{\gcd(2m,n)}-1 \neq 2^{\gcd(m,n)}-1 \\ &\iff \gcd(2m,n) \neq \gcd(m,n) \\ &\iff \nu_2(m)<\nu_2(n). \end{align*} It is now easy to calculate the answer (with casework on $\nu_2(m)$) as $15 \cdot 15+8 \cdot 7+4 \cdot 3+2 \cdot 1=\boxed{295}$.

~Lcz

Remarks

Claim 1 (Olympiad Number Theory Lemma)

If $u,a,$ and $b$ are positive integers such that $u\geq2,$ then $\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.$

There are two proofs to this claim, as shown below.

~MRENTHUSIASM

Claim 1 Proof 1 (Euclidean Algorithm)

If $a=b,$ then $\gcd(a,b)=a=b,$ from which the claim is clearly true.

Otherwise, let $a>b$ without the loss of generality. For all integers $p$ and $q$ such that $p>q>0,$ the Euclidean Algorithm states that \[\gcd(p,q)=\gcd(p-q,q)=\cdots=\gcd(p \ \mathrm{ mod } \ q,q).\] We apply this result repeatedly to reduce the larger number: \[\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^a-1-u^{a-b}\left(u^b-1\right),u^b-1\right)=\gcd\left(u^{a-b}-1,u^b-1\right).\] Continuing, we have \begin{align*} \gcd\left(u^a-1,u^b-1\right)&=\cdots \\ &=\gcd\left(u^{a-b}-1,u^b-1\right) \\ &=\cdots \\ &=\gcd\left(u^{\gcd(a,b)}-1,u^{\gcd(a,b)}-1\right) \\ &=u^{\gcd(a,b)}-1, \end{align*} from which the proof is complete.

~MRENTHUSIASM

Claim 1 Proof 2 (Bézout's Identity)

Let $d=\gcd\left(u^a-1,u^b-1\right).$ It follows that $u^a\equiv1\pmod{d}$ and $u^b\equiv1\pmod{d}.$

By Bézout's Identity, there exist integers $x$ and $y$ such that $ax+by=\gcd(a,b),$ so \[u^{\gcd(a,b)}=u^{ax+by}=\left(u^a\right)^x\left(u^b\right)^y\equiv1\pmod{d},\] from which $u^{\gcd(a,b)}-1\equiv0\pmod{d}.$ We know that $u^{\gcd(a,b)}-1\geq d.$

Next, we notice that \begin{align*} u^a-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{a-\gcd{(a,b)}}+u^{a-2\gcd{(a,b)}}+u^{a-3\gcd{(a,b)}}+\cdots+1\right), \\ u^b-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{b-\gcd{(a,b)}}+u^{b-2\gcd{(a,b)}}+u^{b-3\gcd{(a,b)}}+\cdots+1\right). \end{align*} Since $u^{\gcd(a,b)}-1$ is a common divisor of $u^a-1$ and $u^b-1,$ we conclude that $u^{\gcd(a,b)}-1=d,$ from which the proof is complete.

~MRENTHUSIASM

Claim 2 (GCD Property)

For positive integers $r,s,$ and $t,$ if $\gcd(r,s)=1,$ then $\gcd(r,t)\cdot\gcd(s,t)=\gcd(rs,t).$

As $r$ and $s$ are relatively prime (have no prime divisors in common), this property is intuitive.

~MRENTHUSIASM

See Also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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