X, Y, and Z are three different Prime numbers, the product

Question

X, Y, and Z are three different Prime numbers, the product XYZ is divisible by how many different positive numbers?

A. 4
B. 6
C. 8
D. 9
E. 12

Please describe method.

Bunuel · Accepted Answer

MUST KNOW FOR GMAT: Finding the Number of Factors of an Integer First make prime factorization of an integer n=a^p*b^q*c^r , where a , b , and c are prime factors of n and p , q , and r are their powers. The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1) . NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2 Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors. BACK TO THE ORIGINAL QUESTION: n=xyz (n=x^1y^1z^1) where x , y , and z are different prime factors will have (1+1)(1+1)(1+1)=8 different positive factors including 1 and xyz itself. Answer: C. For more on number properties check: math-number-theory-88376.html Hope it helps.

muralimba · Answer

PRIME # is the # that has only 2 factors: One is 1 and another is the # itself.

infer that

FOR any given # 1 and the # iteself are the definite factors.

Knowing above conepts:

product of X,Y, and Z = XYZ

divisible by 1, xyz, x, y, z, xy,yz,xz ==> total 8

ANSWER "C"

ADDITIONAL INFO.

if question has 5 constants a,b,c,d,e, we do not have to count in the above way

Basically we are selecting, from the product, one constant, set of two constants, set of 3 constants ....and so on set of all the # of constants.

so if 5 varibales are given, total # ways to select is 5C1+5C2+5C3+5C4+5C5 = 5+10+10+5+1 = 31

And answer will be 31+1 =32 (as "1" is a factor for every #)

Regards,
Murali.

Kudos?

tfincham86 · Answer

Thanks for the equation

gettinit · Answer

Thanks Bunuel very helpful, I thought I could do a quick calculation here to find out the number of factors but fell for the trap I guess:

2*3*5=30 so 30 is divisible by 1,30,15,6,5,3,2,10 but in my haste forgot to use 10,and 1 in here. One should just stick to the formula for a sure shot.

muralimba · Answer

Friends,

The approach specified by Bunuel is an efficient, a quick and a standard one. Use the same. Ignore the one specified by me as it is a bit time consuming when compared to that given by Bunuel.

Regards,
Murali.

shahideh · Answer

made a stupid mistake!

Yes the answer is C

1- x- y- z- xy- xz- yz- xyz

kravi2501 · Answer

Ans 8

1
x
y
z
xy
yz
zx
xyz

CrackverbalGMAT · Answer

Let us say X = 2, Y = 3 and Z = 5. Then XYZ = 30 - it is divisible by

1, 2, 3, 5, 6, 10, 15, 30 - 8 different numbers

stonecold · Answer

Since X,Y,Z are different primes=>
Number of factors of X*Y*Z => 2*2*2=> 8
Hence C.

zibisappp · Answer

X, Y, and Z are three different Prime numbers, the product XYZ is divisible by how many different positive numbers?

A. 4
B. 6
C. 8
D. 9
E. 12

Suppose 
X = 2, 
Y = 3 and
Z = 5.

Then XYZ = 30 - it is divisible by

1, 2, 3, 5, 6, 10, 15, 30 - 
so ther are 8 different numbers

answer :-C

Shiv2016 · Answer

Given:x,y,z are prime factors that means they have no factors other than 1 and the no. itself.

Therefore the power of these prime nos. is 1.

The total no. of factors (thus) = 2*2*2=8

ScottTargetTestPrep · Answer

To determine the number of factors of XYZ, or (X^1)(Y^1)(Z^1), we add 1 to each exponent of each unique prime factor and multiply those values together. The result will equal the number of factors of the given number.

Thus, XYZ has (1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 = 8 factors.

Answer: C

gracie · Answer

let p=number of prime numbers=3
total positive factors=2(p+1)=8
C

nigina93 · Answer

I go with C,
each prime number has two factors (1 and itself), so 2*2*2=8

BrentGMATPrepNow · Answer

----ASIDE-------------------------
If the  prime factorization  of N = (p^ a )(q^ b )(r^ c ) . . . (where p, q, r, etc are different prime numbers), then N has a total of ( a +1)( b +1)( c +1)(etc) positive divisors.

Example : 14000 = (2^ 4 )(5^ 3 )(7^ 1 )
So, the number of positive divisors of 14000 = ( 4 +1)( 3 +1)( 1 +1) =(5)(4)(2) = 40
-----------------------------------
Since X, Y and Z are different PRIME numbers, we can say: XYZ = (X^ 1 )(Y^ 1 )(Z^ 1 )
So, the number of positive divisors of XYZ = ( 1 +1)( 1 +1)( 1 +1) =(2)(2)(2) = 8

Answer: C

Cheers, 
Brent

Kinshook · Answer

Given: X, Y, and Z are three different Prime numbers.

Asked: the product XYZ is divisible by how many different positive numbers?

Number of different positive numbers dividing XYZ= 2*2*2=8

IMO C

Posted from my mobile device

sharmashagun770 · Answer

For more on number properties check: https://gmatclub.com/forum/math-number- ... 88376.html I solved it using a different approach. xyz will be divisible by 1, every combination of 2 numbers out of these 3 and the product of all of these. So, the total number of positive numbers that xyz is divisible by 1+3C2 (i.e. 6)+1=8 I have not studied combinations and probability much, so happy to know if my logic is correct or not.

Bunuel · Answer

This is not accurate. First of all, 3C2 = 3!/(2!1!) = 3, not 6. Additionally, the product xyz, where x, y, and z are distinct prime numbers, will be divisible by x, y, and z individually, which is not considered in your solution. Consequently, xyz will have the following factors: 1, x, y, z, xy, xz, yz, and xyz. Using your approach, the correct calculation should be: 1 + 3C1 + 3C2 + 1 = 1 + 3 + 3 + 1 = 8.

Hope it helps.

bumpbot · Answer

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X, Y, and Z are three different Prime numbers, the product

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