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

 
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d &= 2\cdot 3^3\cdot z
 
d &= 2\cdot 3^3\cdot z
 
\end{align*}</cmath>
 
\end{align*}</cmath>
for some positive integers <math>w,x,y,z</math>. Now if <math>3</math> divdes <math>w</math>, then <math>\gcd(a,b)</math> would be at least <math>2^3\cdot 3^2</math> which is too large, hence <math>3</math> does not divide <math>w</math>. Similarly, if <math>2</math> divides <math>z</math>, then <math>\gcd(c,d)</math> would be at least <math>2^2\cdot 3^3</math> which is too large, so <math>2</math> does not divide <math>z</math>. Therefore,
+
for some positive integers <math>w,x,y,z</math>. Now if <math>3</math> divides <math>w</math>, then <math>\gcd(a,b)</math> would be at least <math>2^3\cdot 3^2</math> which is too large, hence <math>3</math> does not divide <math>w</math>. Similarly, if <math>2</math> divides <math>z</math>, then <math>\gcd(c,d)</math> would be at least <math>2^2\cdot 3^3</math> which is too large, so <math>2</math> does not divide <math>z</math>. Therefore,
 
<cmath>\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)</cmath>
 
<cmath>\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)</cmath>
 
where neither <math>2</math> nor <math>3</math> divide <math>\gcd(w,z)</math>. In other words, <math>\gcd(w,z)</math> is divisible only by primes that are at least <math>5</math>. The only possible value of <math>\gcd(a,d)</math> between <math>70</math> and <math>100</math> and which fits this criterion is <math>78=2\cdot3\cdot13</math>, so the answer is <math>\boxed{\textbf{(D) }13}</math>.
 
where neither <math>2</math> nor <math>3</math> divide <math>\gcd(w,z)</math>. In other words, <math>\gcd(w,z)</math> is divisible only by primes that are at least <math>5</math>. The only possible value of <math>\gcd(a,d)</math> between <math>70</math> and <math>100</math> and which fits this criterion is <math>78=2\cdot3\cdot13</math>, so the answer is <math>\boxed{\textbf{(D) }13}</math>.
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Solution by JohnHankock
 
Solution by JohnHankock
  
== Solution 2.1 ==
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== Solution 2.1 (updated with better notation)==
Elaborating on to what Solution 1 stated, we are not able to add any extra factor of 2 or 3 to <math>gcd(a, d)</math> because doing so would later the <math>gcd</math> of <math>(a, b)</math> and <math>(c, d)</math>. This is why:
+
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.
  
The <math>gcd(a, b)</math> is <math>2^3 * 3</math> and the <math>gcd</math> of <math>(c, d)</math> is <math>2 * 3^3</math>. However, the <math>gcd</math> of <math>(b, c) = 2^2 * 3^2</math> (meaning both are divisible by 36). Therefore, <math>a</math> is only divisible by <math>3^1</math> (and no higher power of 3), while <math>d</math> is divisible by only <math>2^1</math> (and no higher power of 2).
+
-Williamgolly
 
 
Thus, the <math>gcd</math> of <math>(a, d)</math> can be expressed in the form <math>2 * 3 * k</math> for which <math>k</math> is a number not divisible by <math>2</math> or <math>3</math>. The only answer choice that satisfies this (and the other condition) is <math>\boxed{\textbf{(D) } 13}</math>.
 
  
 
==Solution 3 (Better notation)==
 
==Solution 3 (Better notation)==
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~Williamgolly
 
~Williamgolly
  
=== Video Solution by Richard Rusczyk ===
+
==Solution 5 (FIREDRAGONMATH16's Solution)==
 +
 
 +
 
 +
Since <math>\gcd (a,b) = 24</math>, <math>a = 24j</math> and <math>b = 24k</math> for some positive integers <math>j,k</math> such that <math>j</math> and <math>k</math> are relatively prime.
 +
 
 +
Similarly , since <math>\gcd (b,c) = 36</math>, we have <math>b = 24k</math> and <math>c=36m</math> with the same criteria. However, since <math>24</math> is not a multiple of <math>36</math>, we must contribute an extra <math>3</math> to <math>b</math> in order to make it a multiple of <math>36</math>. So, <math>k</math> is a multiple of three, and it is relatively prime to <math>m</math>.
 +
 
 +
Finally, <math>\gcd (c,d) = 54</math>, so using the same logic, <math>m</math> is a multiple of <math>3</math> and is relatively prime to <math>n</math> where <math>d = 54n</math>.
 +
 
 +
Since we can't really do anything with these messy expressions, we should try some sample cases of <math>a,b,c</math> and <math>d</math>. Specifically, we let <math>j = 5, 7, 11, 13</math> or <math>17</math>, and see which one works.
 +
 
 +
First we let <math>j= 5</math>. Note that all of these values of <math>j</math> work for the first <math>\gcd</math> expression because they are all not divisible by <math>3</math>.
 +
 
 +
Without the loss of generality, we let <math>k=m=3</math> for all of our sample cases. We can also adjust the value of <math>n</math> in <math>d=54n</math>, since there is no fixed value for <math>\gcd(a,d)</math>; there is only a bound.
 +
 
 +
So we try to make our bound <math>70 < \gcd(a,d) < 100</math> satisfactory. We do so by letting <math>j=n</math>.
 +
 
 +
Testing our first case <math>a=24 \cdot 5</math> and <math>d = 54 \cdot 5</math>, we find that <math>\gcd(a,d) = 30</math>. To simplify our work, we note that <math>\gcd(24,54) = 6</math>, so <math>\gcd(24k, 54k)</math> for all <math>k>1</math> is equal to <math>6k</math>.
 +
 
 +
So now, we can easily find our values of <math>\gcd(a,b)</math>:
 +
 
 +
<cmath>\gcd(24 \cdot 5, 54 \cdot 5) = 6 \cdot 5 = 30</cmath>
 +
 
 +
<cmath>\gcd(24 \cdot 7, 54 \cdot 7) = 6 \cdot 7 = 42</cmath>
 +
 
 +
<cmath>\gcd(24 \cdot 11, 54 \cdot 11) = 6 \cdot 11 = 66</cmath>
 +
 
 +
<cmath>\boxed{\gcd(24 \cdot 13, 54 \cdot 13) = 6 \cdot 13 = 78}</cmath>
 +
 
 +
<cmath>\gcd(24 \cdot 17, 54 \cdot 17) = 6 \cdot 17 = 104</cmath>
 +
 
 +
We can clearly see that only <math>j=n=13</math> is in the bound <math>70 < \gcd(a,d) < 100</math>. So, <math>13</math> must be a divisor of <math>a</math>, which is answer choice <math>\boxed{\textbf{(D)}}</math>.
 +
 
 +
-FIREDRAGONMATH16
 +
 
 +
==Hardness of problem==
 +
 
 +
The ranking of this number theory problem is easy hard(just starting to be hard) or a 7 out of 10. This is because you've got to think through out although it is more of which and easy solution
 +
 
 +
~justin6688
 +
 
 +
== Video Solution by Richard Rusczyk ==
  
 
https://artofproblemsolving.com/videos/amc/2018amc10a/467
 
https://artofproblemsolving.com/videos/amc/2018amc10a/467
  
 
~ dolphin7
 
~ dolphin7
 +
 +
==Video Solution==
 +
https://youtu.be/yjrqINsQP5c
 +
 +
~savannahsolver
 +
 +
== Video Solution (Meta-Solving Technique) ==
 +
https://youtu.be/GmUWIXXf_uk?t=1003
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 07:38, 8 May 2021

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' $2^3 * 3$, that $b$ and $c$ have $2^2 * 3^2$, and that $c$ and $d$ have $3^3 * 2$. Combining $1$ and $2$ yields $b$ has (at a minimum) $2^3 * 3^2$, and thus $a$ has $2^3 * 3$ (and no more powers of $3$ because otherwise $gcd(a,b)$ would be different). In addition, $c$ has $3^3 * 2^2$, and thus $d$ has $3^3 * 2$ (similar to $a$, we see that $d$ cannot have any other powers of $2$). We now assume the simplest scenario, where $a = 2^3 * 3$ and $d = 3^3 * 2$. According to this base case, we have $gcd(a, d) = 2 * 3 = 6$. 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 $6 * 13$ is the only one that works. Therefore the answer is $\boxed{\textbf{(D) } 13}$

Solution by JohnHankock

Solution 2.1 (updated with better notation)

Do casework on $v_2$ and $v_3.$ Notice that we must have $v_3(a) = 1$ and $v_2(d)=1$ and the values of $b,d$ does not matter. Therefore, $\gcd(d,a) = 6k,$ where $k$ is not divisible by $2$ or $3.$ We see that $13$ is the only possible answer.

-Williamgolly

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 $c_3$ similiarly. Now, translate the $lcm$s into the following: \[\min(a_2,b_2)=3\] \[\min(a_3,b_3)=1\] \[\min(b_2,c_2)=2\] \[\min(b_3,c_3)=2\] \[\min(a_2,c_2)=1\] \[\min(a_3,c_3)=3\] .

(Unfinished) ~Rowechen Zhong

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 (FIREDRAGONMATH16's Solution)

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

Hardness of problem

The ranking of this number theory problem is easy hard(just starting to be hard) or a 7 out of 10. This is because you've got to think through out although it is more of which and easy solution

~justin6688

Video Solution by Richard Rusczyk

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

~ dolphin7

Video Solution

https://youtu.be/yjrqINsQP5c

~savannahsolver

Video Solution (Meta-Solving Technique)

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

~ pi_is_3.14

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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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