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yrozenblum
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If p and q are different prime numbers, m and n are integers, and pq 09 Oct 2023, 03:24
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65% (hard)

Question Stats:

66% (02:08) correct 34% (02:25) wrong based on 1054 sessions

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If p and q are different prime numbers, m and n are integers, and pq is a divisor of mp + nq, which of the following must be true?

I. p is a divisor of n.
II. pq is a divisor of mp.
III. p^2 is a divisor of mn.

A. I only
B. II only
C. III only
D. I and II
E. I and III

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2024-01-27_14-10-56.png
2024-01-27_14-10-56.png [ 43.67 KiB | Viewed 15565 times ]
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D
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Bunuel
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If p and q are different prime numbers, m and n are integers, and pq 27 Jan 2024, 03:58
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guddo
If p and q are different prime numbers, m and n are integers, and pq is a divisor of mp + nq, which of the following must be true?

I. p is a divisor of n.
II. pq is a divisor of mp.
III. p^2 is a divisor of mn.

A. I only
B. II only
C. III only
D. I and II
E. I and III

Show Spoiler
Attachment:
2024-01-27_14-10-56.png
Show more

Given that pq is a divisor of mp + nq implies:

    \(\frac{mp + nq}{pq} = integer\)

Splitting the fraction results in:

    \(\frac{mp}{pq} + \frac{nq}{pq} = integer\)

    \(\frac{m}{q} + \frac{n}{p} = integer\)

Since p and q are different prime numbers, we cannot have a case like 3/7 + 11/7 = 2. Therefore, for the sum to be an integer, both \(\frac{m}{q}\) and \(\frac{n}{p}\) must be integers. This means that q is a factor of m and p is a factor of n.

Let's evaluate the options:

I. p is a divisor of n.

As discussed above, this must be true.

II. pq is a divisor of mp.

The above says that \(\frac{mp}{pq}=\frac{m}{q}=integer\), and as discussed above, this must be true.

III. p^2 is a divisor of mn.

However, this is not necessarily true. For example, consider m = 2, n = 3, p = 3, and q = 2.

Answer: D.
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If p and q are different prime numbers, m and n are integers, and pq 19 Apr 2024, 00:48
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Sonia2023
why can't m/q and n/p be fractions and their total is an integer?
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­
\(\frac{mp + nq }{ pq} = k\) in which \(k\) is an integer.­

\(mp + nq = kpq\)

\(mp = kpq - nq\)

\(m = kq - \frac{nq}{p}\) in which \(m\) and \(kq\) are integers, so ­\(\frac{nq}{p}\) must be an integer.

as long as ­\(\frac{nq}{p}\) is an integer, ­\(\frac{n}{p}\) must be an integer because \(p\) and \(q\) are coprime.­
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gmatophobia
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If p and q are different prime numbers, m and n are integers, and pq 09 Oct 2023, 23:38
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yrozenblum
If p and q are different prime numbers, m and n are integers, and pq is a divisor of mp + nq, which of the following must be true?

I. p is a divisor of n.

II. pq is a divisor of mp.

III. p² is a divisor of mn.

A) I only
B) II only
C) III only
D) I and II
E) I and III
Show more

pq is a divisor of mp + nq

\(\frac{mp + nq }{ pq}\) = integer

\(\frac{mp }{ pq} + \frac{nq }{ pq}\) = integer

\(\frac{m }{ q} + \frac{n }{ p}\) = integer

As q and p are different prime numbers, the above statement is possible when \(\frac{m }{ q}\) and \(\frac{n }{ p}\) are integers.

I. p is a divisor of n.

Inference: \(\frac{n}{p}\) = integer

This is correct. For the relationship mentioned in the question to hold true, \(\frac{n}{p}\) must be an integer. We can eliminate B and C.

II. pq is a divisor of mp.

Inference: \(\frac{mp}{pq} = \frac{m}{q} \) = integer

This is correct. For the relationship mentioned in the question to hold true, \(\frac{m}{q}\) must be an integer.

We don't have to check for III as none of the options have all three correct.

Option D
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If p and q are different prime numbers, m and n are integers, and pq 09 Nov 2023, 11:41
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gmatophobia
yrozenblum
If p and q are different prime numbers, m and n are integers, and pq is a divisor of mp + nq, which of the following must be true?

I. p is a divisor of n.

II. pq is a divisor of mp.

III. p² is a divisor of mn.

A) I only
B) II only
C) III only
D) I and II
E) I and III

pq is a divisor of mp + nq

\(\frac{mp + nq }{ pq}\) = integer

\(\frac{mp }{ pq} + \frac{nq }{ pq}\) = integer

\(\frac{m }{ q} + \frac{n }{ p}\) = integer

As q and p are different prime numbers, the above statement is possible when \(\frac{m }{ q}\) and \(\frac{n }{ p}\) are integers.

I. p is a divisor of n.

Inference: \(\frac{n}{p}\) = integer

This is correct. For the relationship mentioned in the question to hold true, \(\frac{n}{p}\) must be an integer. We can eliminate B and C.

II. pq is a divisor of mp.

Inference: \(\frac{mp}{pq} = \frac{m}{q} \) = integer

This is correct. For the relationship mentioned in the question to hold true, \(\frac{m}{q}\) must be an integer.

We don't have to check for III as none of the options have all three correct.

Option D
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gmatophobia - can you explain the highlighted part in bit more detail specifically addressing why can't m/q and n/p be fractions and their total is an integer?

Thanks a lot
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If p and q are different prime numbers, m and n are integers, and pq 08 Dec 2024, 23:20
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Hi Bunuel, are there other problems of this type you can point me to? Got pretty stumped when seeing this problem type

Bunuel
guddo
If p and q are different prime numbers, m and n are integers, and pq is a divisor of mp + nq, which of the following must be true?

I. p is a divisor of n.
II. pq is a divisor of mp.
III. p^2 is a divisor of mn.

A. I only
B. II only
C. III only
D. I and II
E. I and III

Show Spoiler
Attachment:
2024-01-27_14-10-56.png

Given that pq is a divisor of mp + nq implies:


\(\frac{mp + nq}{pq} = integer\)

Splitting the fraction results in:


\(\frac{mp}{pq} + \frac{nq}{pq} = integer\)

\(\frac{m}{q} + \frac{n}{p} = integer\)

Since p and q are different prime numbers, we cannot have a case like 3/7 + 11/7 = 2. Therefore, for the sum to be an integer, both \(\frac{m}{q}\) and \(\frac{n}{p}\) must be integers. This means that q is a factor of m and p is a factor of n.

Let's evaluate the options:

I. p is a divisor of n.

As discussed above, this must be true.

II. pq is a divisor of mp.

The above says that \(\frac{mp}{pq}=\frac{m}{q}=integer\), and as discussed above, this must be true.

III. p^2 is a divisor of mn.

However, this is not necessarily true. For example, consider m = 2, n = 3, p = 3, and q = 2.

Answer: D.
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If p and q are different prime numbers, m and n are integers, and pq 09 Dec 2024, 00:15
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siddysir
Hi Bunuel, are there other problems of this type you can point me to? Got pretty stumped when seeing this problem type

Bunuel
guddo
If p and q are different prime numbers, m and n are integers, and pq is a divisor of mp + nq, which of the following must be true?

I. p is a divisor of n.
II. pq is a divisor of mp.
III. p^2 is a divisor of mn.

A. I only
B. II only
C. III only
D. I and II
E. I and III

Show Spoiler
Attachment:
2024-01-27_14-10-56.png

Given that pq is a divisor of mp + nq implies:


\(\frac{mp + nq}{pq} = integer\)

Splitting the fraction results in:


\(\frac{mp}{pq} + \frac{nq}{pq} = integer\)

\(\frac{m}{q} + \frac{n}{p} = integer\)

Since p and q are different prime numbers, we cannot have a case like 3/7 + 11/7 = 2. Therefore, for the sum to be an integer, both \(\frac{m}{q}\) and \(\frac{n}{p}\) must be integers. This means that q is a factor of m and p is a factor of n.

Let's evaluate the options:

I. p is a divisor of n.

As discussed above, this must be true.

II. pq is a divisor of mp.

The above says that \(\frac{mp}{pq}=\frac{m}{q}=integer\), and as discussed above, this must be true.

III. p^2 is a divisor of mn.

However, this is not necessarily true. For example, consider m = 2, n = 3, p = 3, and q = 2.

Answer: D.
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Multiples and Factors problems: https://gmatclub.com/forum/search.php?s ... tag_id=185

Hope it helps.
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If p and q are different prime numbers, m and n are integers, and pq 06 Jan 2025, 03:51
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yrozenblum
If p and q are different prime numbers, m and n are integers, and pq is a divisor of mp + nq, which of the following must be true?

I. p is a divisor of n.
II. pq is a divisor of mp.
III. p^2 is a divisor of mn.

A. I only
B. II only
C. III only
D. I and II
E. I and III

Show Spoiler
Attachment:
2024-01-27_14-10-56.png
Show more

p and q are different prime numbers (say 2 and 3)

If pq is a divisor of mp + nq, it means p is a divisor of mp + nq and q is also a divisor of mp + nq.

If p is a divisor of mp + nq, we can see that mp is already divisible by p, so nq should also be divisible by p. Since q is a prime number, n MUST be divisible by p.
By the same logic, m MUST be divisible by q.

I. p is a divisor of n - As seen above, it must be true.

II. pq is a divisor of mp - Since m must be divisible by q, mp is divisible by pq. True.

III. p^2 is a divisor of mn - We know that p is a divisor of n for sure but we cannot say anything about p^2.

Answer (D)
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If p and q are different prime numbers, m and n are integers, and pq 10 Jan 2025, 15:42
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This question picking numbers is super simple, I hope I can build this confidence to do it in the exam because in the practice exam I had it wrong.

P=2
M=3
Q=3
N=2

(i) P/N = 2/2 TRUE
(ii) MP/PQ = 6/6 TRUE
(iii) MN/(P**2)=6/9 FALSE
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