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Bunuel
Two numbers are said to be relatively prime if they have no prime factors in common. For example, 8 and 27 are relatively prime because 8 = 2^3 and 27=3^3. If m and n are relatively prime and pn is divisible by m for some integer p, which of the following must also be true?

A. m is a multiple of n
B. m is a multiple of p
C. n is a multiple of p
D. p is a multiple of m
E. p is a multiple of n

it is given that \(m\) & \(n\) are co-prime so \(n\) is not divisible by \(m\)

also \(pn\) is divisible by \(m\)

\(=> \frac{p*n}{m}= integer\), which means that \(p\) has to be a multiple of \(m\) because \(n\) is not divisible by \(m\)

Option D


Hi niks18

i did so:

let m= 3 and n =2

p is some number (any number i choose) so let p= 6

now n*p ---> 2*6 =12

12 is divisible by 3 - is it correct so far? :)

so after this i chose C

what did i do wrong ?

thanks :)

Hi dave13

taking your example only, you have chosen n=2 & p=6.

so now tell me is 2 a multiple of 6 or 6 is a multiple of 2?
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Hello, niks18, thank you for reply and a good question :) Well, 2 is a multiple of 6, because 2*3 = 6 So :? ? :-)
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Hello, niks18, thank you for reply and a good question :) Well, 2 is a multiple of 6, because 2*3 = 6 So :? ? :-)
Hi dave13,
If 2 is a multiple of 6 then 6 has to be a factor of 2 i.e 6 divides 2. Is this correct?
I am posing these questions so that you can find the answers yourself for better understanding. Focus on word “multiple” it is derived from multiplication. So as per your understanding is 2 a result of multiplication of some number with 6? Or is it the other way round?

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Hello, niks18, thank you for reply and a good question :) Well, 2 is a multiple of 6, because 2*3 = 6 So :? ? :-)


Hi..

There are two things..
Multiple and factors...

Multiple...
All integers that come in a certain table..
Here table of 6 is... 6*1=6..6*2=12...6*3=18 and so on
So 6,12, 18 are multiple of 6. So MULTIPLE will be bigger or EQUAL to that number.
. factor..
All integers in whose table that number comes..
2*3=6 means 6 comes in table of 2 and 3, so 2,3 are factors of 6.. factors will be less than or equal to the integer..

So 1,2,3 and 6 are factors of 6 AND 6 is MULTIPLE of 1,2,3, and 6
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Bunuel
Two numbers are said to be relatively prime if they have no prime factors in common. For example, 8 and 27 are relatively prime because 8 = 2^3 and 27=3^3. If m and n are relatively prime and pn is divisible by m for some integer p, which of the following must also be true?

A. m is a multiple of n
B. m is a multiple of p
C. n is a multiple of p
D. p is a multiple of m
E. p is a multiple of n


We have the expression \(\frac{p·n}{m}\) is an integer, and knowing that p is also integer plus that there is no way that m cannot be multiple of n because they share no primes, then p must be multiple of m because no prime interchange occurs between n and m.


AC: D



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Hello chetan2u and niks18, :)

many thanks for taking time to explain. highly appreciated! :) indeed :-)

ok so as per my updated knowledge of multiples and factors i am about to write solution in details:)

Let prime numbers \(m\) and \(n\) be \(2\) and \(3\)respectvely

now, pn is divisible by m for some integer p

So, let \(p\) be \(6\)

(the wording "for some integer p" ) does it for 6 ?

\(\frac{pn}{m}\) =\(\frac{6*3}{2}\)= \(9\)

(the wording "for some integer p" ) does it mean for 6 ?

if yes, than how can 6 be multiple of m (2 in this case) because it contradicts Chetan`s explanation =

6 multiple of 2 means to get 6 i multiply 2 by 3

but it can also be: 6 multiple of 3: to get 6 i multiply 3 by 2 which means p is a multiple of n as well :? :)

i googled:
For examples, 2, 4, 6, 8, and 10 are multiples of 2. To get these numbers, you multiplied 2 by 1, 2, 3, 4, and 5, which are integers. A multiple of a number is that number multiplied by an integer.

agained confused something :? :-)
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Bunuel
Two numbers are said to be relatively prime if they have no prime factors in common. For example, 8 and 27 are relatively prime because 8 = 2^3 and 27=3^3. If m and n are relatively prime and pn is divisible by m for some integer p, which of the following must also be true?

A. m is a multiple of n
B. m is a multiple of p
C. n is a multiple of p
D. p is a multiple of m
E. p is a multiple of n

Here, the question says m & n are not multiple of each other as they are relatively prime. So, for the statement in passage \(\frac{p*n}{m}\) to be integer \(\frac{p}{m}\) should be integer. Hence, D.
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dave13
Hello chetan2u and niks18, :)

many thanks for taking time to explain. highly appreciated! :) indeed :-)

ok so as per my updated knowledge of multiples and factors i am about to write solution in details:)

Let prime numbers \(m\) and \(n\) be \(2\) and \(3\)respectvely

now, pn is divisible by m for some integer p

So, let \(p\) be \(6\)

(the wording "for some integer p" ) does it for 6 ?

\(\frac{pn}{m}\) =\(\frac{6*3}{2}\)= \(9\)

(the wording "for some integer p" ) does it mean for 6 ?

if yes, than how can 6 be multiple of m (2 in this case) because it contradicts Chetan`s explanation =

6 multiple of 2 means to get 6 i multiply 2 by 3

but it can also be: 6 multiple of 3: to get 6 i multiply 3 by 2 which means p is a multiple of n as well :? :)

i googled:
For examples, 2, 4, 6, 8, and 10 are multiples of 2. To get these numbers, you multiplied 2 by 1, 2, 3, 4, and 5, which are integers. A multiple of a number is that number multiplied by an integer.

agained confused something :? :-)

Hi dave13
6 is a multiple of 2
6=2*3 and 6 is also multiple of 3
So in this case p is multiple of m & n both.
Now read the question carefully, it says “Must be True”. So if I choose p=8, then p is a multiple of 2 but not 3. Hence pn is always divisible by m but not by n. This is the reason why option E is incorrect.
But I hope you understood why option C is incorrect!

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