abhirup1711 wrote:Three is the largest number that can be divided evenly into 27 and the positive integer X, while 10 is the largest number that can be divided equally into both 100 and X. Which of the following is the largest possible number that could be divided into X, 2 and 100?

30
70
210
300
700

Please help. The language of the problem is a bit strange.

Yeah, I'm not too crazy about the language either.
Which of the following is the largest possible number that could be divided into X, 2 and 100?

Only 2 positive integers (1 and 2) divide into 2.
Since all of the answer choices are greater than 2, I don't see how any of them can divide INTO 2.

What's the source of the question?

Cheers,
Brent

I did some searching, and it looks like the question should read as follows:

Three is the largest number that an be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100.
A) 30
B) 70
C) 210
D) 300
E) 700

Some background information
For questions involving factors (aka "divisors"), we can say:
If k is a divisor of N, then k is "hiding" within the prime factorization of N
Examples:
3 is a divisor of 24 <--> 24 = (2)(2)(2)(3)
5 is a divisor of 70 <--> 70 = (2)(5)(7)
8 is a divisor of 56 <--> 56 = (2)(2)(2)(7)

Okay, onto the solution.

3 is the largest number that an be divided evenly into 27 and the positive integer x.
In other words, 3 is the Greatest Common Divisor (GCD) of 27 and x
27 = (3)(3)(3)
If the GCD of 27 and x is 3, then the prime factorization of x has exactly one 3 (but no more than one 3)
At this point, we know that the prime factorization of x looks like this:
x = (3)(?)(?)(?)(?)(?)

Aside: We have a free video about finding the Greatest Common Divisor of a pair of values: https://www.gmatprepnow.com/module/gmat- ... ies?id=833


10 is the largest number that can be divided equally into both 100 and X
In other words, 10 is the Greatest Common Divisor (GCD) of 100 and x
100 = (2)(2)(5)(5)
If the GCD of 100 and x is 10 (aka 2 times 5), then the prime factorization of x has exactly one 2 and one 5
We now know that the prime factorization of x looks like this:
x = (2)(5)(?)(?)(?)(?)

When we combine both pieces of information, we know that the prime factorization of x looks like this:
x = (2)(3)(5)(?)(?)(?)(?)
Since (2)(3)(5) = 30, we know that x is a multiple of 30. So, we can already eliminate B and E.

Since we're looking for the greatest possible value of x, let's start with the biggest answer choices.

D) 300
x cannot equal 300. Here's why.
300 = (2)(2)(3)(5)(5), and we already concluded that the prime factorization of x has exactly one 2 and one 5 (otherwise, the GCD of x and 100 will be greater than 10)
Since the prime factorization of 300 has more than one 2 and one 3, we can ELIMINATE D

C) 210
x can equal 210
210 = (2)(3)(5)(7), and this adheres to our conclusions that the prime factorization of x has exactly one 2, and exactly one 2 and one 5


Answer: C

Cheers,
Brent