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[m]N = m^4n^2p^3[/m]

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KarunMendiratta
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[m]N = m^4n^2p^3[/m] [#permalink] New post   Updated on: 21 Mar 2021, 22:50
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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 \(N\)
\(60\)


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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C

Originally posted by KarunMendiratta on 10 Mar 2021, 20:20.
Last edited by Carcass on 21 Mar 2021, 22:50, edited 1 time in total.
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eldorado21
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Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   11 Mar 2021, 16:46
Some more details in the question are needed,

is m = n = p ? Is m,n,p greater than 1?
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Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   11 Mar 2021, 23:35
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eldorado21 wrote:
Some more details in the question are needed,

is m = n = p ? Is m,n,p greater than 1?


Thanks, I have made the changes.
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Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   14 Mar 2021, 01:03
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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] New post   21 Mar 2021, 16:46
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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] New post   21 Mar 2021, 21:54
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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] New post   28 May 2021, 19:55
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] New post   29 May 2021, 01:29
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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] New post   29 May 2021, 04:30
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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] New post   19 Jul 2022, 11:08
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] New post   19 Jul 2022, 11:14
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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] New post   19 Jul 2022, 11:18
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] New post   19 Jul 2022, 11:25
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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

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
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Re: [m]N = m^4n^2p^3[/m] [#permalink] New post   19 Jul 2022, 11:27
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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] New post   07 Mar 2024, 04:41
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