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M04-31

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M04-31  [#permalink]

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New post 16 Sep 2014, 00:23
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A perfect number is defined as one for which the sum of all the unique factors less the number itself is equal to the number. For instance, 6 is a perfect number, because the factors of 6 (apart from 6 itself) are 1, 2 and 3, and \(1 + 2 + 3 = 6\). Which of the following is also a perfect number?

A. 12
B. 20
C. 28
D. 48
E. 60

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New post 16 Sep 2014, 00:23
Official Solution:

A perfect number is defined as one for which the sum of all the unique factors less the number itself is equal to the number. For instance, 6 is a perfect number, because the factors of 6 (apart from 6 itself) are 1, 2 and 3, and \(1 + 2 + 3 = 6\). Which of the following is also a perfect number?

A. 12
B. 20
C. 28
D. 48
E. 60

\(28 = 1+2+4+7+14\)


Answer: C
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New post 04 Oct 2014, 10:57
How we have calculated Factor of 28 ? 4 and 14 are not prime .. is it correct to show factor as non prime number ?
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New post 04 Oct 2014, 11:11
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rahulgmat2014 wrote:
How we have calculated Factor of 28 ? 4 and 14 are not prime .. is it correct to show factor as non prime number ?


Factor of an integer is not necessarily a prime number.

A divisor of an integer \(n\), also called a factor of \(n\), is an integer which evenly divides \(n\) without leaving a remainder. In general, it is said \(m\) is a factor of \(n\), for non-zero integers \(m\) and \(n\), if there exists an integer \(k\) such that \(n = km\).

So, the factors of 28 are 1, 2, 4, 7, 14, and 28.

Theory on Number Properties: math-number-theory-88376.html
Divisibility tips: divisibility-multiples-factors-tips-and-hints-174998.html?hilit=divisibility%20tips

DS Divisibility/Multiples/Factors questions to practice: search.php?search_id=tag&tag_id=354
PS Divisibility/Multiples/Factors questions to practice: search.php?search_id=tag&tag_id=185[/textarea]

Hope it helps.
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Re: M04-31  [#permalink]

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New post 16 Nov 2014, 16:55
So when it says "unique factors' - that does not equal prime?
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New post 17 Nov 2014, 02:01
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New post 09 Sep 2016, 07:23
Bunuel wrote:
Official Solution:


\(28 = 1+2+4+7+14\)


Answer: C


Why 12 is incorrect 1+2+3+6 = 12
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New post 09 Sep 2016, 08:04
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New post 03 Mar 2018, 15:56
Bunuel, any way to narrow down some options, so it wont be 100% try and error?
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New post 05 Aug 2018, 06:47
Is there a faster way to find ALL the factors of a number?
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New post 05 Aug 2018, 07:14
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imran1994 wrote:
Is there a faster way to find ALL the factors of a number?


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.

2. Properties of Integers




5. Divisibility/Multiples/Factors



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Re: M04-31  [#permalink]

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New post 30 Nov 2018, 08:23
Bunuel wrote:
A perfect number is defined as one for which the sum of all the unique factors less the number itself is equal to the number. For instance, 6 is a perfect number, because the factors of 6 (apart from 6 itself) are 1, 2 and 3, and \(1 + 2 + 3 = 6\). Which of the following is also a perfect number?

A. 12
B. 20
C. 28
D. 48
E. 60

This is question is simple to calculate for all , To calculate sum of factors except the number , we calculate sum of all factors - the number.

Sum of all factors of a number is given by if a number is P= a^n * b^m * c^k. Sum of all factors of P = (1 + a ..... a^n) (1+b+b^2 .... b^m )(1+c + c^2+....c^k)

so 28 will have (1+ 2 + 4) (1+ 7) = 56 sum of factors ,excluding number will be 28, so the perfect number.
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Re: M04-31   [#permalink] 30 Nov 2018, 08:23
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