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

Revision as of 08:59, 12 September 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

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 1.1

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 2 (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

@@ Line 14: / Line 14: @@
 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).
-Thus, the <math>gcd</math> of <math>(a, d)</math> can be expressed in the form <math>2 * 3 * k</math> for which k is a number not divisible by 2 or 3. The only answer choice that satisfies this 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)==

2018 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2018 AMC 10A Problems/Problem 22"

Revision as of 08:59, 12 September 2018

Contents

Solution 1

Solution 1.1

Solution 2 (Better notation)

See Also