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I am a a student and I am having difficulty with answering this question. I keep getting the answer wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.

n is a number. 100 is the LCM of 20 and n. Work out two different possible values for n.

n = ______ n = ______

I did this: Hcf of 100:1,2,4,5,10,20,25,50,100

I don't know what to do next?

Thank you and help would be appreciated

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    Welcome. The best way to get help here is to show your work so far. Please edit your question, putting in the steps you took and the wrong answer to arrived at. – 2017-01-21
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    Did you find *one* possible value for $n$? – 2017-01-21
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    No well not yet – 2017-01-21
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    Since the LCM is a common *multiple*, 100 must be divisible by both $20$ (it is) and $n$. Try out a few divisors of $100$ for $n$ and see which work. – 2017-01-21
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    Hint: look at the prime factorization of 20 – 2017-01-21
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    5 is divisible by 20 and 100. It's also a prime factor – 2017-01-21
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    20 is divisible by 5, not the other way around, but their lcm is not 100 – 2017-01-21
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    Is it possible if I could get the answer with a step by step solution so I can understand in the future. I don't quite understand. – 2017-01-21
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    Hint: $n$ can not be a divisor of $20$ (otherwise the LCM would be 20), but it must be a divisor of $100$. This leaves only a few numbers to try out. – 2017-01-21
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    It's 25 and 50 because 1,2,4,5 ,10 and 20 are divisors of 20 so it must be 25 and 50. Thank you Ethan Bolker and dxiv for guiding me through these questions. – 2017-01-21

2 Answers 2

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First of all, get the divisors of 20:

20 is divisible by 1,2,4,5,10,20

Then let's look at the divisors of 100:

100 is divisible by 1,2,4,5,10,20,25,50, 100.

The only possible solutions are 25, 50 and 100 because n can not be a divisor of 20.

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    Thank you for showing me how to do the question. Now I won't be stuck on these type of questions. – 2017-01-21
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$20=2^2×5$

Suppose $n$ is of the form
$2^a×5^b$ where $a \ge 2; b\ge1$

Then $2^a×5^b=100$ $\implies a=2,b=2 \implies n=100$

Suppose $n$ is of the form
$5^b$ where $b\ge1$

Then $2^2×5^b=100$ $\implies b=2 \implies n=25$

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    Thank you for showing me how to do these types of questions. – 2017-01-21