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If n is a prime number between 0 and 100, how many positive divisors

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If n is a prime number between 0 and 100, how many positive divisors  [#permalink]

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New post 20 Apr 2016, 00:03
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Re: If n is a prime number between 0 and 100, how many positive divisors  [#permalink]

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New post 20 Apr 2016, 01:37
I'd say the answer is 1 because every divisor is just a multiple of the original prime?
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Re: If n is a prime number between 0 and 100, how many positive divisors  [#permalink]

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New post 20 Apr 2016, 01:51
3
mathiaskeul wrote:
I'd say the answer is 1 because every divisor is just a multiple of the original prime?



Hi,
we are basically looking it factors of\(n^3,\) where n is a PRIME number..
number of factors \(= (3+1) = 4.\).
we can also write them down.
1, n, \(n^2\), and \(n^3\)
so ans will be 4
D
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If n is a prime number between 0 and 100, how many positive divisors  [#permalink]

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New post 20 Apr 2016, 01:57
chetan2u wrote:
mathiaskeul wrote:
I'd say the answer is 1 because every divisor is just a multiple of the original prime?



Hi,
we are basically looking it factors of\(n^3,\) where n is a PRIME number..
number of factors \(= (3+1) = 4.\).
we can also write them down.
1, n, \(n^2\), and \(n^3\)
so ans will be 4
D


Hi chetan. Could you explain the rule for this? Is it only for prime numbers?

Thank you!
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If n is a prime number between 0 and 100, how many positive divisors  [#permalink]

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New post 20 Apr 2016, 02:03
mathiaskeul wrote:
chetan2u wrote:
mathiaskeul wrote:
I'd say the answer is 1 because every divisor is just a multiple of the original prime?



Hi,
we are basically looking it factors of\(n^3,\) where n is a PRIME number..
number of factors \(= (3+1) = 4.\).
we can also write them down.
1, n, \(n^2\), and \(n^3\)
so ans will be 4
D


Hi chetan. Could you explain the rule for this? Is it only for prime numbers?

Thank you!


Hi,

when ever you are looking for the number of factors/positive divisors, get the integer in its simplest form/scientific notation ..
for example say 36..
\(36 = 2^2*3^2\).. ans will be (2+1)(2+1) = 3*3=9..
\(120 = 2^3*3*5\).. ans will be (3+1)(1+1)(1+1) = 4*2*2=16..

say a Q further asks you ODD factor of 120..
just drop the even 2-- (3+1)(1+1)(1+1) .. ans will be (1+1)(1+1)=2*2=4..
Even factors will be Total - ODD = 16-4=12..
May help somewhere
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If n is a prime number between 0 and 100, how many positive divisors  [#permalink]

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New post 20 Apr 2016, 02:05
mathiaskeul wrote:
chetan2u wrote:
mathiaskeul wrote:
I'd say the answer is 1 because every divisor is just a multiple of the original prime?



Hi,
we are basically looking it factors of\(n^3,\) where n is a PRIME number..
number of factors \(= (3+1) = 4.\).
we can also write them down.
1, n, \(n^2\), and \(n^3\)
so ans will be 4
D


Hi chetan. Could you explain the rule for this? Is it only for prime numbers?

Thank you!


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.

Check for more here: math-number-theory-88376.html
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Re: If n is a prime number between 0 and 100, how many positive divisors  [#permalink]

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New post 20 Apr 2016, 12:45
Two ways to solve this one
Pick has 2 => 8 has 4 factors hence D
next way to solve this up is using the formula => if K=P^a * Q^b * R^c where P,Q,R are primes ; then the number of divisors of K are => a+1 * b+1 * c+1


Hit that D option .
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Re: If n is a prime number between 0 and 100, how many positive divisors  [#permalink]

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New post 02 Jul 2018, 09:02
Bunuel wrote:
If n is a prime number between 0 and 100, how many positive divisors does n^3 have?

A. 1
B. 2
C. 3
D. 4
E. 5


If let n = 2, then n^3 = 8, which has divisors of 1, 2, 4, and 8, so n^3 has 4 positive divisors. This concept holds for any prime number n^3. It is always/only divisible by 1, n, n^2, and n^3.

Answer: D
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Re: If n is a prime number between 0 and 100, how many positive divisors &nbs [#permalink] 02 Jul 2018, 09:02
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