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Is there any easier or algorithmic method to solve this problem?

There are two numbers, $n$ and $6$. The least common multiple of $n$ and $6$ is $24$. Find $n$.

The way I do this is by expressing $6$ and $24$ into their prime factorisations.

$6=2\cdot3$ and $24=2^3\cdot3$. By simply observing and do simple reasoning, we can see that $n$ can be either $8$ or $24$. Right?

I know that if we are given its $\operatorname{gcd}$, then we can use $n\times6=\operatorname{lcm}(n,6)\cdot \operatorname{gcd}(n,6)$. But we are not given its $\operatorname{gcd}$ here.

If we try to explain this to a student who just learned about $\operatorname{lcm}$, how can we explain it in a simplest manner? Thanks.

5xum
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user71346
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3 Answers3

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Hint:

$\gcd(n,6)=1,2,3 \text{ or } 6$ since it must be a divisor of $6$.

Try each in turn to see which give answers which work with the original question and which do not.

Henry
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first of all express 24 in prime factorization.now we know that lcm of any two given numbers is the product of highest powers of prime factors. 24=2^3.3 in it highest power of 2 is 3.If we express 6 in prime factorization highest power of 2 is 1 (6=2.3) hence n must contain highest power of 2 i.e 2^3 must be a factor of n. Highest power of 3 is 1 and 6 has 3 as a factor so n may or may not contain 3. Hence n may be 2^3=8 or may be 2^3.3=24.

user273057
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So, $n$ must divide $\displaystyle24\implies n$ is of the form $2^a3^b$ where integers $0\le a\le3,0\le b\le1$

So, $\operatorname{lcm}\displaystyle(n,6)=2^{\text{max}(a,1)}\cdot3^{\text{max}(b,1)}=24=2^3\cdot3^1$

So, $b$ can be $0,1$

Hope you can take it home from here?

5xum
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lab bhattacharjee
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