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Bunuel
If x is divisible by 12 and 18, which of the following must divide evenly into x?

I. 6
II. 36
III. 72

A. I only
B. I and II
C. II​ and III
D. II​ only
E. I, II, and III

Answer B
First number divisible by both 12 and 18=36. Now, 72 does not evenly divide into 36. Therefore, only 6 and 36 satisfy
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Bunuel
If x is divisible by 12 and 18, which of the following must divide evenly into x?

I. 6
II. 36
III. 72

A. I only
B. I and II
C. II​ and III
D. II​ only
E. I, II, and III

\(12\) = \(2^2\) x \(3\)

\(18\)= \(3^2\) x \(2\)

The least value of X will be 6 {\(2^1\) x \(3^1\)}

Thus the number will be divisible by 6

Further -

Least common multiple of 12 & 18 is 36 { Divisible by 18 as well as 12}

Thus 36 will also divide X

Answer will be both I and II

This question basically tests the application of HCF and LCM
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If X can be divided evenly by 12 and 18 , it must be divided by the LCM of the 2 numbers
LCM of 12 and 18 = 36

So if hypothetically a number can be divided by 12, 18 and 36 , it will also be divided by 6.

Lets check the third number 72.

What if number to be divided turns out to be 36. Can it be divided by 72 ? NO

Hence, Correct Option is B
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If x is divisible by 12 and 18, which of the following must divide evenly into x?

I. 6
II. 36
III. 72

A. I only
B. I and II
C. II​ and III
D. II​ only
E. I, II, and III

12 = 2^2 x 3 and 18 = 2 x 3^2. If a number is divisible by both 12 and 18, the number has at least two 2's and two 3's in its prime factorization.

So we can write x = 2^2 x 3^2 x m (m is an integer) or x = 36m

This implies x is always divisible by 6. (2 x 3). I is always true. Eliminate C and D.
II is always true as well. As x=36m; x is always divisible by 36. Eliminate A.

III not always true. Now x=36m. if m=1; x=36. this is not divisible by 72. Hence not always true.

Eliminate E.

this leaves B.
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General expression for x=> 36*p for some integer p
so C is the answer as 36 and 72 are possible but 6 is not.
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