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Re: [m]N = m^4n^2p^3[/m] [#permalink]
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KarunMendiratta wrote:
N is an integer which can be written in it's factor form as: \(N = m^4n^2p^3\), Where, \(1 < m < n < p\)

Quantity A
Quantity B
Number of positive factors of \(N\)
\(60\)


OA Explanation

If \(N = (a^x)(b^y)(c^z)....\)
Then, Total number of factors = \((x + 1)(y + 1)(z + 1)....\)

Col. A: Number of factors = \((4 + 1)(2 + 1)(3 + 1) = (5)(3)(4) = 60\)
Col. B: \(60\)

Hence, option C
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Re: [m]N = m^4n^2p^3[/m] [#permalink]
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Carcass

KarunMendiratta


Help here!

Ans must be D,

m n and p can be compund numbers, what if m is 11, n is 26 and p is 77?

Kind regards!
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Re: [m]N = m^4n^2p^3[/m] [#permalink]
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FCOCGALVAN

The questions says Factor Form!

\((11)(27)(77)\) would be written as \((3^3)(7)(11^2)\)
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Re: [m]N = m^4n^2p^3[/m] [#permalink]
KarunMendiratta wrote:
KarunMendiratta wrote:
N is an integer which can be written in it's factor form as: \(N = m^4n^2p^3\), Where, \(1 < m < n < p\)

Quantity A
Quantity B
Number of positive factors of \(N\)
\(60\)


OA Explanation

If \(N = (a^x)(b^y)(c^z)....\)
Then, Total number of factors = \((x + 1)(y + 1)(z + 1)....\)

Col. A: Number of factors = \((4 + 1)(2 + 1)(3 + 1) = (5)(3)(4) = 60\)
Col. B: \(60\)

Hence, option C


Your calculated factor is total no of factor sir, but Quantity A ask for no of positive factor which must be 60/2 =30. Hence, quantity B must be greater.
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Re: [m]N = m^4n^2p^3[/m] [#permalink]
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kumarneupane4344 wrote:
KarunMendiratta wrote:
KarunMendiratta wrote:
N is an integer which can be written in it's factor form as: \(N = m^4n^2p^3\), Where, \(1 < m < n < p\)

Quantity A
Quantity B
Number of positive factors of \(N\)
\(60\)


OA Explanation

If \(N = (a^x)(b^y)(c^z)....\)
Then, Total number of factors = \((x + 1)(y + 1)(z + 1)....\)

Col. A: Number of factors = \((4 + 1)(2 + 1)(3 + 1) = (5)(3)(4) = 60\)
Col. B: \(60\)

Hence, option C


Your calculated factor is total no of factor sir, but Quantity A ask for no of positive factor which must be 60/2 =30. Hence, quantity B must be greater.


kumarneupane4344
Dear Sir, I have only calculated the positive factors of N here, there are 60 more factors of N which are negative.

Example:
Let N = 8 = \(2^3\)

Total number of positive factors = 3 + 1 = 4 (1, 2, 4, and 8)
Number of negative factors = 4 (-1, -2, -4, and -8)

Hence, Total number of factors of N = 4 + 4 = 8
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Re: [m]N = m^4n^2p^3[/m] [#permalink]
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Thank you KarunMendiratta sir, I have a misconception about this formula that it gives total no of factors. NOw I am clear sir. Many thanks, Stay safe!!
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Re: [m]N = m^4n^2p^3[/m] [#permalink]
Carcass Can factor form be assumed as "Prime Factor"? I am still unsure why D doesn't hold true

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Re: [m]N = m^4n^2p^3[/m] [#permalink]
Expert Reply
I am not sure what you meant

However, when you perform the factorization they are all primes

usually we do have composite numbers but also just prime numbers that can be divided only by 1 and themselves
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Re: [m]N = m^4n^2p^3[/m] [#permalink]
Carcass

I meant if a factored form is given in the question stem does it always have to be Prime raised to exponents.

From your reply, it seems it always is

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[m]N = m^4n^2p^3[/m] [#permalink]
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AppK wrote:
Carcass

I meant if a factored form is given in the question stem does it always have to be Prime raised to exponents.

From your reply, it seems it always is

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\(8^2 = 64\) and neither 8 nor 64 are primes

But at its core \(8^2\) is just \(2^3 * 2^3 = 2^6\) and 2 is prime
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Re: [m]N = m^4n^2p^3[/m] [#permalink]
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Carcass wrote:
AppK wrote:
Carcass

I meant if a factored form is given in the question stem does it always have to be Prime raised to exponents.

From your reply, it seems it always is

Posted from my mobile device


\(8^2 = 64\) and neither 8 nor 64 are primes

But at its core \(8^2\) is just \(2^3 * 2^3 = 2^6\) and 2 is prime


In fact, when you see question like this they are always factorization problems and they are always problems prime-related
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Re: [m]N = m^4n^2p^3[/m] [#permalink]
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