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X, Y, and Z are three different Prime numbers, the product [#permalink]
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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.
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Re: xyz prime [#permalink]
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gettinit wrote: X, Y, and Z are three different Prime numbers, the product XYZ is divisible by how many different positive numbers?
4 6 8 9 12
Please describe method. 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?



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Re: xyz prime [#permalink]
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gettinit wrote: X, Y, and Z are three different Prime numbers, the product XYZ is divisible by how many different positive numbers?
4 6 8 9 12
Please describe method. MUST KNOW FOR GMAT: Finding the Number of Factors of an IntegerFirst 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: mathnumbertheory88376.htmlHope it helps.
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Re: xyz prime [#permalink]
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10 Dec 2010, 07:52



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Re: xyz prime [#permalink]
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10 Dec 2010, 12:55
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.



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Re: xyz prime [#permalink]
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11 Dec 2010, 00:14
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.



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Re: xyz prime [#permalink]
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20 Oct 2011, 07:16
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made a stupid mistake!
Yes the answer is C
1 x y z xy xz yz xyz



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Re: xyz prime [#permalink]
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Ans 8
1 x y z xy yz zx xyz



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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]
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21 Jan 2014, 23:33
gettinit wrote: 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. 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
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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]
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16 Jan 2017, 00:03
Since X,Y,Z are different primes=> Number of factors of X*Y*Z => 2*2*2=> 8 Hence C.
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X, Y, and Z are three different Prime numbers, the product [#permalink]
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20 Jan 2017, 01:51
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



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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]
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03 Apr 2017, 11:01
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
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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]
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06 Apr 2017, 10:03
gettinit wrote: 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 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
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Re: X, Y, and Z are three different Prime numbers, the product [#permalink]
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07 Apr 2017, 13:53
gettinit wrote: 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. let p=number of prime numbers=3 total positive factors=2(p+1)=8 C




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