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# How many prime factors does the number X have?

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3074
How many prime factors does the number X have?  [#permalink]

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Updated on: 13 Aug 2018, 02:23
00:00

Difficulty:

65% (hard)

Question Stats:

51% (01:31) correct 49% (01:39) wrong based on 194 sessions

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e-GMAT Question:

How many prime factors does the number X have?

1) X is divisible by 53
2) X is divisible by 4 distinct integers.

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient

This is

Question 8 of The e-GMAT Number Properties Marathon

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Question 9 of the Marathon

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Originally posted by EgmatQuantExpert on 27 Feb 2018, 10:13.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:23, edited 2 times in total.
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Re: How many prime factors does the number X have?  [#permalink]

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27 Feb 2018, 12:45
1
EgmatQuantExpert wrote:

Question:

How many prime factors does the number X have?

1) X is divisible by 53
2) X is divisible by 4 distinct integers.

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient

if x is divisible by 53, it can have other primes as well

insufficient

x is divisible by 4 distinct integers

1, x , and 2 others

now here the other 2 must be prime

14 = 1 2 7 14
15 = 1 3 5 15
sufficient

but if

x = cube of a prime i.e 2^3 , which has 4 factors

we get 1 prime

hence insufficient

combining x can be cube of a prime or multiple of 2 squares of prime

(E) imo
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Re: How many prime factors does the number X have?  [#permalink]

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27 Feb 2018, 22:43
3
EgmatQuantExpert wrote:

Question:

How many prime factors does the number X have?

1) X is divisible by 53
2) X is divisible by 4 distinct integers.

A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D) EACH statement ALONE is sufficient.
E) Statements (1) and (2) TOGETHER are NOT sufficient

Statement 1

If X is divisible by 53, then X could just be '53' (in which case only 1 prime factor, 53) or X could be '53*2 = 106' (in which case 2 prime factors, 2 and 53).
So we cannot be sure about the number of prime factors of X. Not Sufficient.

Statement 2

X is divisible by 4 distinct integers, this means X has total 4 factors. Now any number having 4 factors in total, could have two kinds of prime factorisation:

a) X = p^3. So X could be cube of a prime number. In this case X will have total 4 factors, but only 1 prime factor (p).
b) X = p1 * p2. So X could be a product of two distinct prime numbers. In this case also X will have total 4 factors, but there are 2 prime factors (p1 and p2).

So we cannot be sure about number of prime factors of X. Not Sufficient.

Combining the statements, we know X is divisible by 53 and X has 4 total factors.

So if X looks like p^3, then X = 53^3. In this case X has only 1 prime factor 53.
But if X looks like p1 * p2, then X = 53*p2 (product of 53 with any other prime number). In this case X has 2 prime factors 53 and p2.

So we still cannot be sure, as there are two possibilities. Not Sufficient.

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Joined: 04 Jan 2015
Posts: 3074
Re: How many prime factors does the number X have?  [#permalink]

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28 Feb 2018, 11:25

Solution:

Step 1: Analyse Statement 1:
$$X$$ is divisible by $$53$$.
• The given statement tells us that $$X$$ is a multiple of $$53$$.
o So, from this statement we can write $$X$$ in the form of $$53k$$, where $$k$$is any positive integer.
• To find the number of prime factors of $$X=53k$$, we need to have information on the value of $$k$$.
As we do not have any information on $$k$$,
Statement 1 alone is NOT sufficient to answer the question.
Hence, we can eliminate answer choices A and D.
Step 2: Analyse Statement 2:
$$X$$ is divisible by $$4$$ distinct integers
• The given statement tells us that $$X$$ has $$4$$ factors.
o It can be of form $$P1 * P2$$, where $$P1$$and $$P2$$ are the prime factors of the number.
o Or, it can be of the form $$P1^3$$, where $$P1$$ is the prime factor of the number.
• Hence, it can have either $$1$$or $$2$$ prime factors.
Statement 2 alone is NOT sufficient to answer the question.
Step 3: Combine both Statements:
• From the first statement, we got $$X=53k$$
• From the second statement we got $$X$$has $$4$$ factors
o Combining the statements, we may encounter two possible cases:
 It can be of form $$53 * P2$$, where $$P2$$ is a prime factor of the number.
 Or, it can be of form$$53^3$$.
• Hence, it can have either $$1$$ or $$2$$ prime factors.

By combining both statements we did not get a unique answer.
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Re: How many prime factors does the number X have?  [#permalink]

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06 May 2019, 09:41
Hi - wondering why is this a 700 question? Don't think there is anything out of the ordinary one needs to spot? Or is that on the basis of the responses given?
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Re: How many prime factors does the number X have?  [#permalink]

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20 May 2019, 05:07
Hatakekakashi wrote:
EgmatQuantExpert wrote:

Question:

but if

x = cube of a prime i.e 2^3 , which has 4 factors

we get 1 prime

hence insufficient

if x = cube of prime, does x have 4 DISTINCT integers??? I think the answer should be B.
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How many prime factors does the number X have?  [#permalink]

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01 Oct 2019, 05:58
EgmatQuantExpert wrote:

e-GMAT Question:

How many prime factors does the number X have?

1) X is divisible by 53
2) X is divisible by 4 distinct integers.

1) X is divisible by 53: insufic.
if X=53y and y=1 then $$pf(X)=53^1=(1+1)=2$$
if X=53y and y=2 then $$pf(X)=53^1•2^1=(1+1)(1+1)=4$$

2) X is divisible by 4 distinct integers: insufic.
if factors(X)={1,2,3,5} then $$pf(X)=2^1•3^1•5^1=(2)(2)(2)=8$$
if factors(X)={2,3,5,7} then $$pf(X)=2^1•3^1•5^1•7^1=(2)(2)(2)(2)=16$$

(1&2) insufic.
if factors(X)={1,2,3,53} then $$pf(X)=2^1•3^1•53^1=(2)(2)(2)=8$$
if factors(X)={2,3,7,53} then $$pf(X)=2^1•3^1•7^1•53^1=(2)(2)(2)(2)=16$$

How many prime factors does the number X have?   [#permalink] 01 Oct 2019, 05:58
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