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sushom101
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How many positive integers k are there such that 100k is a factor of ( 29 Oct 2005, 13:55
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65% (hard)

Question Stats:

52% (01:22) correct 48% (01:16) wrong based on 668 sessions

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How many positive integers k are there such that 100k is a factor of (2^2)(3)(5^3)?

(A) None
(B) One
(C) Two
(D) Three
(E) Four
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E
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londonluddite
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How many positive integers k are there such that 100k is a factor of ( 07 Nov 2006, 13:19
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100 = 2^2*5^2

so k could be:
1 to give 2^2*5^2*1
3 to give 2^2*3*5^2
5 to give 2^2*5^3
15 to give 2^2*3*5^3
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gmatnerdforever
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How many positive integers k are there such that 100k is a factor of ( 26 Sep 2020, 17:03
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taking out 100 fom the expersion 2^2 * 3 * 5^3 we are left with 3*5*1 so the folowing number can take up the velues of k = 3,5,1,15 therefore the answer is 4
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How many positive integers k are there such that 100k is a factor of ( 26 Sep 2020, 20:30
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hkhanexam
how many positive integers k are there such that 100k is a factor of (2^2)(3)(5^3)?

A) none
B) 1
C) 2
D) 3
E) 4

I found this in another discussion saying that "This discussion does not meet community quality standards. It has been retired."

Please help!
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For any number, N broken into prime factors and their powers in the form \(N = a^p * b^q*c^r\) and so on, then the total number of factors n = (p + 1) * (q + 1) * (r + 1) and so on..

We want to know, how many values of k are possible, if 100k is a factor of the given number i.e we need to find out how many factors of \((2^2)(3)(5^3)\) end with 2 zeros.

To get a number ending with a zero, we need a 2 and a 5, and for 2 zeroes, we need 2 twos and 2 fives.

Taking out 2 two and 2 fives, we get \(2^2 * 5^2[3^1 * 5^1]\)

Now by the counting rule for factors, we count the number of factors for the numbers in the bracket i.e \(3^1 * 5^1\)

The number of factors = (1 + 1) * (1 + 1) = 4


Option E

Arun Kumar
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How many positive integers k are there such that 100k is a factor of ( 02 Oct 2020, 12:45
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CrackVerbalGMAT
hkhanexam
how many positive integers k are there such that 100k is a factor of (2^2)(3)(5^3)?

A) none
B) 1
C) 2
D) 3
E) 4

I found this in another discussion saying that "This discussion does not meet community quality standards. It has been retired."

Please help!

For any number, N broken into prime factors and their powers in the form \(N = a^p * b^q*c^r\) and so on, then the total number of factors n = (p + 1) * (q + 1) * (r + 1) and so on..

We want to know, how many values of k are possible, if 100k is a factor of the given number i.e we need to find out how many factors of \((2^2)(3)(5^3)\) end with 2 zeros.

To get a number ending with a zero, we need a 2 and a 5, and for 2 zeroes, we need 2 twos and 2 fives.

Taking out 2 two and 2 fives, we get \(2^2 * 5^2[3^1 * 5^1]\)

Now by the counting rule for factors, we count the number of factors for the numbers in the bracket i.e \(3^1 * 5^1\)

The number of factors = (1 + 1) * (1 + 1) = 4


Option E

Arun Kumar
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I don't understand the highlighted portion, I know the counting rule to count factors, but not understanding why we're counting the remaining 5 and 3 factors.
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How many positive integers k are there such that 100k is a factor of ( 04 Mar 2025, 10:53
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