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# Greatest Common Divisor | AMC-10A, 2018 | Problem 22

Try this beautiful problem from ALGEBRA: Greatest Common Divisor AMC-10A, 2018. You may use sequential hints to solve the problem

Try this beautiful problem from Algebra based on Greatest Common Divisor from AMC 10A, 2018, Problem 22.

## Greatest Common Divisor – AMC-10A, 2018- Problem 22

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 $a$?

• $5$
• $7$
• $13$

### Key Concepts

Number theory

Gcd

Divisior

Answer: $13$

AMC-10A (2018) Problem 22

Pre College Mathematics

## Try with Hints

TO find the divisor of $a$ at first we have to find the value of $a$.can you find the value of $a$?

Given that $\gcd(a, b)=24$, $\gcd(b, c)=36$, $\gcd(c, d)=54$, and $70<\gcd(d, a)<100$

so we can say $a=24 \times$ some integer and $b=24 \times$ some another integer (according to gcd rules)

similarly for the others c & d…..

now if we can find out the value of $\gcd(d, a)$ then we may use the condition $70<\gcd(d, a)<100$

Can you now finish the problem ……….

so we may say that $gcd(a, b)$ 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).

can you finish the problem……..

so we can say that $gcd$ of $(a, d)$ can be expressed in the form $2 \times 3 \times$ some positve integer and now $k$ is a number not divisible by $2$ or $3$. so from the given numbes it will be $13$ because $2 \times 3 \times k$ must lie $70<\gcd (d, a)<100$. so the required ans is $13$

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