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Difference between revisions of "2018 AMC 10A Problems/Problem 22"

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Solution by JohnHankock
 
Solution by JohnHankock
  
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== Solution 2.1 (updated with better notation)==
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==Solution 3 (Better notation)==
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Do casework on <math>v_2</math> and <math>v_3.</math>  Notice that we must have <math>v_3(a) = 1</math> and <math>v_2(d)=1</math> and the values of <math>b,d</math> does not matter.  Therefore, <math>\gcd(d,a) = 6k,</math> where <math>k</math> is not divisible by <math>2</math> or <math>3.</math>  We see that <math>13</math> is the only possible answer.
 
  
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-Williamgolly
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First off, note that <math>24</math>, <math>36</math>, and <math>54</math> are all of the form <math>2^x\times3^y</math>. The prime factorizations are <math>2^3\times 3^1</math>, <math>2^2\times 3^2</math> and <math>2^1\times 3^3</math>, respectively. Now, let <math>a_2</math> and <math>a_3</math> be the number of times <math>2</math> and <math>3</math> go into <math>a</math>, respectively. Define <math>b_2</math>, <math>b_3</math>, <math>c_2</math>, and <math>c_3</math> similarly. Now, translate the <math>lcm</math>s into the following:
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<cmath>1) \min(a_2,b_2)=3</cmath> <cmath>2) \min(a_3,b_3)=1</cmath> <cmath>3) \min(b_2,c_2)=2</cmath> <cmath>4) \min(b_3,c_3)=2</cmath> <cmath>5) \min(c_2,d_2)=1</cmath> <cmath>6) \min(c_3,d_3)=3</cmath>
  
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==Solution 3 (Better notation)==
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From <math>4)</math>, we see that <math>b_3 \geq 2</math>, thus from <math>2)</math>, <math>a_3 = 1</math>. Similarly, from <math>3)</math>, <math>c_2 \geq 2</math>, thus from <math>5)</math>, <math>d_2 = 1</math>.
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Note also that <math>d_3 \geq 3</math> and <math>a_2 \geq 3</math>. Therefore <cmath>\min(a_2, d_2) = 1</cmath> <cmath>\min(a_3, d_3) = 1</cmath> Thus, <math>\gcd(a, d) = 2 \times 3 \times k</math> for some <math>k</math> having no factors of <math>2</math> or <math>3</math>.
  
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First off, note that <math>24</math>, <math>36</math>, and <math>54</math> are all of the form <math>2^x\times3^y</math>. The prime factorizations are <math>2^3\times 3^1</math>, <math>2^2\times 3^2</math> and <math>2^1\times 3^3</math>, respectively. Now, let <math>a_2</math> and <math>a_3</math> be the number of times <math>2</math> and <math>3</math> go into <math>a</math>,respectively. Define <math>b_2</math>, <math>b_3</math>, <math>c_2</math>, and <math>c_3</math> similiarly. Now, translate the <math>lcm</math>s into the following:
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Since <math>70 < \gcd(a, d) < 100</math>, the only values for k are <math>12, 13, 14, 15, 16</math>, but all have either factors of <math>2</math> or <math>3</math>, except <math>\boxed{\textbf{(D)} 13}</math>. And we're done.  
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<cmath>\min(a_2,b_2)=3</cmath> <cmath>\min(a_3,b_3)=1</cmath> <cmath>\min(b_2,c_2)=2</cmath> <cmath>\min(b_3,c_3)=2</cmath> <cmath>\min(a_2,c_2)=1</cmath> <cmath>\min(a_3,c_3)=3</cmath> .
 
  
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(Unfinished)
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~Rowechen Zhong ~MannsNotHot
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~Rowechen Zhong
 
  
 
==Solution 4 (Fastest)==
 
==Solution 4 (Fastest)==
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~savannahsolver
 
~savannahsolver
  
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== Video Solution (Meta-Solving Technique) ==
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== Video Solution by OmegaLearn (Meta-Solving Technique) ==
 
https://youtu.be/GmUWIXXf_uk?t=1003
 
https://youtu.be/GmUWIXXf_uk?t=1003
  

Latest revision as of 03:51, 6 August 2023

Problem

Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$, $\gcd(b, c)=36$, $\gcd(c, d)=54$, and $70<\gcd(d, a)<100$. Which of the following must be a divisor of $a$?

$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

Solution 1

The GCD information tells us that $24$ divides $a$, both $24$ and $36$ divide $b$, both $36$ and $54$ divide $c$, and $54$ divides $d$. Note that we have the prime factorizations: \begin{align*} 24 &= 2^3\cdot 3,\\ 36 &= 2^2\cdot 3^2,\\ 54 &= 2\cdot 3^3. \end{align*}

Hence we have \begin{align*} a &= 2^3\cdot 3\cdot w\\ b &= 2^3\cdot 3^2\cdot x\\ c &= 2^2\cdot 3^3\cdot y\\ d &= 2\cdot 3^3\cdot z \end{align*} for some positive integers $w,x,y,z$. Now if $3$ divides $w$, then $\gcd(a,b)$ would be at least $2^3\cdot 3^2$ which is too large, hence $3$ does not divide $w$. Similarly, if $2$ divides $z$, then $\gcd(c,d)$ would be at least $2^2\cdot 3^3$ which is too large, so $2$ does not divide $z$. Therefore, \[\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)\] where neither $2$ nor $3$ divide $\gcd(w,z)$. In other words, $\gcd(w,z)$ is divisible only by primes that are at least $5$. The only possible value of $\gcd(a,d)$ between $70$ and $100$ and which fits this criterion is $78=2\cdot3\cdot13$, so the answer is $\boxed{\textbf{(D) }13}$.

Solution 2

We can say that $a$ and $b$ 'have' $2^3 \cdot 3$, that $b$ and $c$ have $2^2 \cdot 3^2$, and that $c$ and $d$ have $3^3 \cdot 2$. Combining $1$ and $2$ yields $b$ has (at a minimum) $2^3 \cdot 3^2$, and thus $a$ has $2^3 \cdot 3$ (and no more powers of $3$ because otherwise $\gcd(a,b)$ would be different). In addition, $c$ has $3^3 \cdot 2^2$, and thus $d$ has $3^3 \cdot 2$ (similar to $a$, we see that $d$ cannot have any other powers of $2$). We now assume the simplest scenario, where $a = 2^3 \cdot 3$ and $d = 3^3 \cdot 2$. According to this base case, we have $\gcd(a, d) = 2 \cdot 3 = 6$. We want an extra factor between the two such that this number is between $70$ and $100$, and this new factor cannot be divisible by $2$ or $3$. Checking through, we see that $6 \cdot 13$ is the only one that works. Therefore the answer is $\boxed{\textbf{(D) } 13}$

Solution by JohnHankock

Solution 3 (Better notation)

First off, note that $24$, $36$, and $54$ are all of the form $2^x\times3^y$. The prime factorizations are $2^3\times 3^1$, $2^2\times 3^2$ and $2^1\times 3^3$, respectively. Now, let $a_2$ and $a_3$ be the number of times $2$ and $3$ go into $a$, respectively. Define $b_2$, $b_3$, $c_2$, and $c_3$ similarly. Now, translate the $lcm$s into the following: \[1) \min(a_2,b_2)=3\] \[2) \min(a_3,b_3)=1\] \[3) \min(b_2,c_2)=2\] \[4) \min(b_3,c_3)=2\] \[5) \min(c_2,d_2)=1\] \[6) \min(c_3,d_3)=3\]

From $4)$, we see that $b_3 \geq 2$, thus from $2)$, $a_3 = 1$. Similarly, from $3)$, $c_2 \geq 2$, thus from $5)$, $d_2 = 1$.

Note also that $d_3 \geq 3$ and $a_2 \geq 3$. Therefore \[\min(a_2, d_2) = 1\] \[\min(a_3, d_3) = 1\] Thus, $\gcd(a, d) = 2 \times 3 \times k$ for some $k$ having no factors of $2$ or $3$.

Since $70 < \gcd(a, d) < 100$, the only values for k are $12, 13, 14, 15, 16$, but all have either factors of $2$ or $3$, except $\boxed{\textbf{(D)} 13}$. And we're done.

~Rowechen Zhong ~MannsNotHot

Solution 4 (Fastest)

Notice that $\gcd (a,b,c,d)=\gcd(\gcd(a,b),\gcd(b,c),\gcd(c,d))=\gcd(24,36,54)=6$, so $\gcd(d,a)$ must be a multiple of $6$. The only answer choice that gives a value between $70$ and $100$ when multiplied by $6$ is $\boxed{\textbf{(D) } 13}$. - mathleticguyyy + einstein

In the case where there can be 2 possible answers, we can do casework on $\gcd(d,a)$ ~Williamgolly

Solution 5

Since $\gcd (a,b) = 24$, $a = 24j$ and $b = 24k$ for some positive integers $j,k$ such that $j$ and $k$ are relatively prime.

Similarly , since $\gcd (b,c) = 36$, we have $b = 24k$ and $c=36m$ with the same criteria. However, since $24$ is not a multiple of $36$, we must contribute an extra $3$ to $b$ in order to make it a multiple of $36$. So, $k$ is a multiple of three, and it is relatively prime to $m$.

Finally, $\gcd (c,d) = 54$, so using the same logic, $m$ is a multiple of $3$ and is relatively prime to $n$ where $d = 54n$.

Since we can't really do anything with these messy expressions, we should try some sample cases of $a,b,c$ and $d$. Specifically, we let $j = 5, 7, 11, 13$ or $17$, and see which one works.

First we let $j= 5$. Note that all of these values of $j$ work for the first $\gcd$ expression because they are all not divisible by $3$.

Without the loss of generality, we let $k=m=3$ for all of our sample cases. We can also adjust the value of $n$ in $d=54n$, since there is no fixed value for $\gcd(a,d)$; there is only a bound.

So we try to make our bound $70 < \gcd(a,d) < 100$ satisfactory. We do so by letting $j=n$.

Testing our first case $a=24 \cdot 5$ and $d = 54 \cdot 5$, we find that $\gcd(a,d) = 30$. To simplify our work, we note that $\gcd(24,54) = 6$, so $\gcd(24k, 54k)$ for all $k>1$ is equal to $6k$.

So now, we can easily find our values of $\gcd(a,b)$:

\[\gcd(24 \cdot 5, 54 \cdot 5) = 6 \cdot 5 = 30\]

\[\gcd(24 \cdot 7, 54 \cdot 7) = 6 \cdot 7 = 42\]

\[\gcd(24 \cdot 11, 54 \cdot 11) = 6 \cdot 11 = 66\]

\[\boxed{\gcd(24 \cdot 13, 54 \cdot 13) = 6 \cdot 13 = 78}\]

\[\gcd(24 \cdot 17, 54 \cdot 17) = 6 \cdot 17 = 104\]

We can clearly see that only $j=n=13$ is in the bound $70 < \gcd(a,d) < 100$. So, $13$ must be a divisor of $a$, which is answer choice $\boxed{\textbf{(D)}}$.

-FIREDRAGONMATH16

Solution 6

[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z, W; real R; path quad; //Variable Definitions R = 1; X = R*dir(45); Y = R*dir(135); Z = R*dir(-135); W = R*dir(-45); quad = X--Y--Z--W--cycle; //Diagram draw(quad); label("$b$",X,NE); label("$a=2^3 \cdot 3 \cdot p$",Y,NW); label("$d=2 \cdot 3^3 \cdot q$",Z,SW); label("$c$",W,SE); label("$2^3 \cdot 3$",X--Y); label("$2^2 \cdot 3^2$",X--W); label("$2 \cdot 3^3$",Z--W); label("$2 \cdot 3 \cdot k$",Z--Y); [/asy]

The relationship of $a$, $b$, $c$, and $d$ is shown in the above diagram. $gcd(a,d)=2 \cdot 3 \cdot k$. $70 < 6k < 100$, $12 \le k \le 16$, $k=\boxed{\textbf{(D) }13}$

Note that $gcd(b,c)=36$ is not required to solve the problem.

~isabelchen


Solution 7 (Easier version of Solution 1)

Just as in solution $1$, we prime factorize $a, b, c$ and $d$ to observe that

$a=2^3\cdot{3}\cdot{w}$

$b=2^3\cdot{3^2}\cdot{x}$

$c=2^2\cdot{3^3}\cdot{y}$

$d=2\cdot{3^3}\cdot{z}.$

Substituting these expressions for $a$ and $d$ into the final given,

$70<\text{gcd}(2\cdot{3^3}\cdot{z}, 2^3\cdot{3}\cdot{w})<100.$

The greatest common divisor of these two numbers is already $6$. If $k$ is what we wish to multiply $6$ by to obtain the gcd of these two numbers, then

$70<6k<100$. Testing the answer choices, only $13$ works for $k$ (in order for the compound inequality to hold). so our gcd is $78$, which means that $\boxed{\textbf{(D) }13}$ must divide $a$.

-Benedict T (countmath1)

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2018amc10a/467

~ dolphin7

Video Solution

https://youtu.be/yjrqINsQP5c

~savannahsolver

Video Solution by OmegaLearn (Meta-Solving Technique)

https://youtu.be/GmUWIXXf_uk?t=1003

~ pi_is_3.14

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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