The highest common factor of x and 15 is 3, the highest common factor

let us evaluate
We know HCF is the least common factor of the two numbers

s.no Factors of X factor of given number HCF Indication/inference
1 X 15=3*5 3 X has 3 as a factor but not 5
2 X 140=2*2*5*7 2*7 X has 2&7 as a factor but not 4&5
3 X 660=2*2*3*5*11 ???

We know from s.no 1 and 2 X has factors 3,2 & 7 but not 5 &4 but we do not have any information about 11 , since we have been asked to find max possible HCF let us consider 11 as a factor of X

so max possible HCF in the given condition of x and 660 is 3*2*11=66 i.e answer E

1/ The highest common factor between X and 15 is 3. It means that X is multiple of 3 and not multiple of 5
2/ The highest common factor between X and 140 is 14. We know that 140 = 2^2*5*7. It means that X is multiple of 2 and 7, but not multiple of 2^2.
From 1 and 2 => X = 2*3*7*A (A is an integer)
We know that 660 = 2^2*3*5*11
The highest common factor between X and 660 should be the product of 2*3*B (B should be 1 or 11).

=> the possible highest common factor between X and 660 is 2*3*11 = 66

Answer E

Bunuel - can you share links of more such questions? Thanks in advance.

 
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Check in Multiples and Factors.

Hope this helps.

Why should B be 1 or 11?
How are we considering 11 as a possibility?

From the given data in question, we know that x cannot have \(2^2\), it also cannot have 5. And what x has is a 3, a 2 and a 7 and possibly an unknown prime k which isn't 2 or 5.

When you break down 660 to its prime factor form: \(660 => 3*11*5*2^2\)

What x can have in common with 660 is of course a 3, and a 2.

When you look at options do we see a 6? No we don't => there exists an unknown prime k.

Do we have strong grounds to eliminate options A, B, C and D? Yes we have by comparing prime factors of x and 660.

For 66, we get 3*2*11 as its prime factors, do we have any reason to eliminate this? No we don't, we can very well assume that 11 can be the unknown prime k in our dataset as it does not go against any inferences we have made from the question.

Hence we consider 11 as a possibility. Hope it helps.