Find the number of factors of the number p

SonalSinha803

Note that P<100 and P=x^12

Given: p<100

To calculate the number of factors of p, we need the prime factors of p and the exponents of the prime factors.

1. The number p has odd number of factors.

this implies that the number p is a perfect square. So, it can be 1,4,9,16,25,36,49,64,81
Not Sufficient

2. the number p can be expressed as x^12,y^3,z^6

As p is less than 100, so we can have only one value for p
1 ^12= 1^3 = 1^6
p= 1

Correct Answer = B

1. The number p has odd number of factors. >>>> If a number has odd factors that means its a perfect square in this case P can be 4,16,25,36,49,64,81 all are <100 , so St1 Not Sufficient.
2. The number p can be expressed as \(x^{12}\), \(y^{3}\) or \(z^{6}\) where x, y, and z are positive integers. >>>>> This implies:

a. P = \(x^{12}\) < 100 then x can be only 1 because \(2^{12}\) > 100.
b. P = \(y^{3}\) < 100 then y can be 1,2,3 or 4.
c. P = \(z^{6}\) < 100 then z can be 1 or 2.

From a,b and c its clear that the only factor common among them is 1 and because P = \(1^{12}\) < 100 St2 is sufficient.

Answer: B.

I) for odd no of factors power of number has to even
Eg 2^4 gives me 5 factors
2^2 gives me 3 factors
Different answers so not sufficient

II)2^ 12 is not possible which is smallest integer so only value which satisfy this is 1.

So B gives definite answer.

Posted from my mobile device

The correct answer is option B

The solution is fairly straight forward - From the question p<100.
Consider Option 1 - It is clearly not helpful. (Advanced : This does help us understand that p is made up of even powers of prime factors)
Now Option 2 - The number can be expressed as \(x^{12}\), where x is a positive integer. Let's consider x=2. \(2^{12}\) = \(64^{2}\) , clearly >100. So x <2, hence the only option is 1.

Hence option B is the final answer.

B is the correct answer.

S1: p=odd; this is not sufficient as there can be many numbers which are less than 100 and have many odd factors
S2: if P can be written in the form of x^12, y^3 or z^6 then supposing
x=2 will give 2^12>100;
x=1 will give 1^12<100; x can have value 1
y=1 will give 1^3<100
y=2 will give 2^3<100; this will hold true for y=3 & 4; so y can have values : 1, 2, 3, 4
Z=1 will give 1^6<100
Z=2 will give 2^6<100
Z=3 will give 3^6>100; thus z can have values 1,2

So combining x, y, z we get p=1 hence B is sufficient.

For statement 1,
Only square number have odd number of factors.
9 can be written as 3^2. so 3 factors.
36 can be written as 2^2 * 3^2. so 3*3 factor.

But unique answer cannot be obtained.

For statement 2,
If we take the smallest number greater than 1 i.e. 2, we get 2^12 >100
So the number should be 1.

We can get a unique answer here so the B should be IMO.


Please give Kudos if the answer is correct or the solution is helpful to you!!!!!!

Though the question doesn't say if x, y, z are prime, we can arrive at x=1 as no other value satisfies the conditions p=\(x^12\) and p<100 simultaneously.
How is the answer E then.

egmat
Please recheck the OA, it should be B.



Posted from my mobile device

The OA should be B, we can easily confirm as X^12 is provided.

Thanks for attempting this one, everybody. My apologies for a typo in the question. The correct question (edited in the main post) is


Find the number of factors of the number p, if p is a positive integer less than 100.

1. The number p has odd number of factors.

2. The number p can be expressed as \(x^{2}\), \(y^{3}\) or \(z^{6}\) where x, y, and z are positive 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.

Although there was a typo (apologies again), some great work by all of you. rkgstyle - Kudos for a great explanation! Please PM me to get your GMATClub tests!

Good day everyone,

I have a question concerning the answer being E. I understand that with statement 2 we might arrive to 2 answers which would be 1 or 64. But for me - as I am understanding the way the question is asked -, the answer ''1'' is not valid. Indeed, the statement 2 cleary says that x, y and z are positive integers in plural. That means, therefore, that the integers x, y and z are 3 different integers and only the answer ''64'' satisfy this condition.

Am I wrong for thinking like that?

Thx for your help

Hello

The question just says: x, y, z are positive integers. That doesnt necessarily mean that they are distinct, because the word 'distinct' has not been used in the question.

From statement 2 x,y,z are positive integers and p can be expressed as x^2,y^3,z^6.The only integer 64 satisfies this condition.
Hence I suppose answer as B.
Please correct me

x =1,2 both satisfied statement 2 hence it is not sufficient.

See if x = 1, x^2,x^3,x^6 =1 which is less than 100.
If x =2, max value is x^6=64, less than 100.
Hence x can be 1 or 2, unique value can't be found. Hence it is not sufficient.

B is not right answer.

Answer shall be E.




Posted from my mobile device

I feel if the terms are given as x, y and z , they are bound to be distinct terms.
This leads us to the fact that P != 1 , and P= 2 , can be the only solution.
IMO B is the correct choice here.

P.S. : I see some folks have used x^12 , while I see x^2 in the question. Can anyone clarify?