Is X divisible by 15?

1. When X is divided by 10, the result is an integer

2. X is a multiple of 30An important aspect of any data sufficiency question is to apply to the

FOIN checks to ensure we coverall the different types of values that X may assume.

Many a time, when answering data sufficiency questions, we overlook the fact that the X may be ZERO, negative, fraction or an irrational number.

In order to establish that X is divisible by 15, we have to establish BOTH :-

a) X is an integer (Note : 0 is considered divisible by any integer)

b) X is divisible by BOTH 5 and 3.Here are the FOIN checks :-

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F - Can the number be a

Fraction ?

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0 - Can the number be

zero ?

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I - Can the number be an

Integer or an

irrational number ?

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N - Can the number be

negative ?

Lets consider the answer options and apply the FOIN

a) When n is divided by 10, the result is an integerF - Can X be a fraction ?

No. Any fraction divided by 10 leads to yet another fraction

0 - Can X be zero.

Yes it can.. ZERO Divided by anything is an integer.

I - Can X be an integer or an irrational number....

No...X has to be an integer , as any irrational number divided by a

integer can only lead to another irrational number

N - Can X be negative -

Yes it can. But this fact doesnt bother us too much as we dont care about whether X is negative

or positive, just the fact whether X is a multiple of 15.

So from 1) we have established that X is an integer and that X is divisible by 10.But wait, if X is

divisible by 10, X is divisible by

all PRIME factors of 10. This implies that X is divisible by

2,5 (all PRIME factors of 10). We have

not been able to conclusively establish that X is

also divisible by 3.

For e.g : X may be 100 or 300. Both 100 and 300 are divisible by 10, but only 300 is divisible by 3.

HENCE 1) on its own is INSUFFICIENT...

Lets take 2

X^2 is a multiple of 30When we apply the FOIN test, we find out that X may be an irrational number.

X can be SQRT(30).

SQRT(30) * SQRT(30) = 30, which is divisible by 30.

Hence lets not bother with further checks as this statement is clearly insufficient on its own. Remember we need to

prove that X is an integer.

Let's for a second evaluate the statement

" X ^ 2" is multiple of 30. Let's understand the implication of this statement.

For this discussion, let's momentarily assume that

X is an integer.

"

X ^ 2" is multiple of 30 ======>

X^2 has 30 as its factor==>

X^2 has all factors of 30 as its factor

Therefore

X^2 has 2,3,5 as its PRIME factors.

Think about this for a minute, we get X^2 when we multiple X * X.

Hence if we look at the PRIME factors of X^2, we should see each PRIME factor of X occurring 2 times.

Look at an example

Lets say X = 6 . The PRIME factors of X are 2,3.

X * X = 36

Factors of 36 should include two occurrences of 2 and two occurrences of 3.

which makes sense as 36 = 2 *2 * 3*3.

So what this implies is that if (X ^2) has 2,3,5 as its factors, then X should also have 2,3,5 as its PRIME factorsIf we knew for sure that X was an integer, this statement would have been sufficient on its own. However, we dont know for

sure that X is an integer.

Hence (2) on its own is not sufficient.

1 + 2 ==> X is an integer and that X has 2,3,5 as its factors.

Hence (C)

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