Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
Line 4: Line 4:
  
 
== Solution 1 ==
 
== Solution 1 ==
 +
 +
The gcd information tells us that 24 divides <math>a</math>, both 24 and 36 divide <math>b</math>, both 36 and 54 divide <math>c</math>, and 54 divides <math>d</math>. Note that we have the prime factorizations:
 +
<cmath>\begin{align*}
 +
24 &= 2^3\cdot 3,\\
 +
36 &= 2^2\cdot 3^2,\\
 +
54 &= 2\cdot 3^3.
 +
\end{align*}</cmath>
 +
 +
Hence we can write
 +
<cmath>\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*}</cmath>
 +
for some positive integers <math>w,x,y,z</math>. Now if 3 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 3 does not divide <math>w</math>. Similarly, if 2 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 2 does not divide <math>z</math>. Therefore,
 +
<cmath>\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)</cmath>
 +
where neither 2 nor 3 divide <math>\gcd(w,z)</math>. In other words, <math>\gcd(w,z)</math> are divisible only by primes that are at least 5. The only number of this form between 70 and 100 is <math>78=2\cdot3\cdot13</math>, so the answer is <math>\boxed{\textbf{(D) }13}</math>.
 +
 +
== Solution 2 ==
  
 
We can say that <math>a</math> and <math>b</math> 'have' <math>2^3 * 3</math>, that <math>b</math> and <math>c</math> have <math>2^2 * 3^2</math>, and that <math>c</math> and <math>d</math> have <math>3^3 * 2</math>. Combining <math>1</math> and <math>2</math> yields <math>b</math> has (at a minimum) <math>2^3 * 3^2</math>, and thus <math>a</math> has <math>2^3 * 3</math> (and no more powers of <math>3</math> because otherwise <math>gcd(a,b)</math> would be different). In addition, <math>c</math> has <math>3^3 * 2^2</math>, and thus <math>d</math> has <math>3^3 * 2</math> (similar to <math>a</math>, we see that <math>d</math> cannot have any other powers of <math>2</math>). We now assume the simplest scenario, where <math>a = 2^3 * 3</math> and <math>d = 3^3 * 2</math>. According to this base case, we have <math>gcd(a, d) = 2 * 3 = 6</math>. We want an extra factor between the two such that this number is between <math>70</math> and <math>100</math>, and this new factor cannot be divisible by <math>2</math> or <math>3</math>. Checking through, we see that <math>6 * 13</math> is the only one that works. Therefore the answer is <math>\boxed{\textbf{(D) } 13}</math>
 
We can say that <math>a</math> and <math>b</math> 'have' <math>2^3 * 3</math>, that <math>b</math> and <math>c</math> have <math>2^2 * 3^2</math>, and that <math>c</math> and <math>d</math> have <math>3^3 * 2</math>. Combining <math>1</math> and <math>2</math> yields <math>b</math> has (at a minimum) <math>2^3 * 3^2</math>, and thus <math>a</math> has <math>2^3 * 3</math> (and no more powers of <math>3</math> because otherwise <math>gcd(a,b)</math> would be different). In addition, <math>c</math> has <math>3^3 * 2^2</math>, and thus <math>d</math> has <math>3^3 * 2</math> (similar to <math>a</math>, we see that <math>d</math> cannot have any other powers of <math>2</math>). We now assume the simplest scenario, where <math>a = 2^3 * 3</math> and <math>d = 3^3 * 2</math>. According to this base case, we have <math>gcd(a, d) = 2 * 3 = 6</math>. We want an extra factor between the two such that this number is between <math>70</math> and <math>100</math>, and this new factor cannot be divisible by <math>2</math> or <math>3</math>. Checking through, we see that <math>6 * 13</math> is the only one that works. Therefore the answer is <math>\boxed{\textbf{(D) } 13}</math>
Line 9: Line 29:
 
Solution by JohnHankock
 
Solution by JohnHankock
  
−
== Solution 1.1 ==
+
== Solution 2.1 ==
 
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:  
 
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:  
  
Line 16: Line 36:
 
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>.
 
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 2 (Better notation)==
+
==Solution 3 (Better notation)==
  
 
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:  
 
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:  

Revision as of 23:35, 10 December 2018

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 can write \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 divdes $w$, 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, $\gcd(w,z)$ are divisible only by primes that are at least 5. The only number of this form between 70 and 100 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

Elaborating on to what Solution 1 stated, we are not able to add any extra factor of 2 or 3 to $gcd(a, d)$ because doing so would later the $gcd$ of $(a, b)$ and $(c, d)$. This is why:

The $gcd(a, b)$ is $2^3 * 3$ and the $gcd$ of $(c, d)$ is $2 * 3^3$. However, the $gcd$ of $(b, c) = 2^2 * 3^2$ (meaning both are divisible by 36). Therefore, $a$ is only divisible by $3^1$ (and no higher power of 3), while $d$ is divisible by only $2^1$ (and no higher power of 2).

Thus, the $gcd$ of $(a, d)$ can be expressed in the form $2 * 3 * k$ for which $k$ is a number not divisible by $2$ or $3$. The only answer choice that satisfies this (and the other condition) is $\boxed{\textbf{(D) } 13}$.

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_22&oldid=99369"