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

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== Solution ==
 
== Solution ==
  
We can say that a and b 'have' <math>2^3 * 3</math>, 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 the 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 'worst case 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 necessarily becomes between 70 and 100. Checking through, we see that 6 * 13 -> D is the only one that works.
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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>. Checking through, we see that <math>6 * 13</math> is the only one that works. Therefore the answer is <math>\boxed{D}</math>
  
 
Solution by JohnHankock
 
Solution by JohnHankock
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==See Also==
 +
{{AMC10 box|year=2018|ab=A|num-b=21|num-a=23}}

Revision as of 18:54, 8 February 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

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 $100$. Checking through, we see that $6 * 13$ is the only one that works. Therefore the answer is $\boxed{D}$

Solution by JohnHankock

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 10 Problems and Solutions
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