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

## Check the Answer

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 \(13\)

## Other useful links

- https://www.cheenta.com/linear-equations-amc-8-2007problem-20/
- https://www.youtube.com/watch?v=5fWkdSs5PZk