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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

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 \(13\)

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