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The integers m and p are such that 2<m<p, and m is not a

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The integers m and p are such that 2<m<p, and m is not a [#permalink]

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06 May 2010, 11:30
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The integers m and p are such that 2<m<p, and m is not a factor of p. If r is the remainder when p is divided by m, is r>1?

(1) The greatest common factor of m and p is 2.

(2) The least common multiple of m and p is 30.
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Mar 2012, 01:48, edited 1 time in total.
Edited the question and added the OA

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06 May 2010, 12:01
I think it's C.

2 < m < p , m is not a factor of p

St. 1 : GCF is 2 - Insufficient --> m,p values could be 6,10 or 8,12 or 4,10
St. 2 : LCM is 30 - Insufficient --> m,p values could be 3,10 or 5,6 or 6,10

St. 1 & 2 : Sufficient --> only value is 6,10 --> we can get the value of r.

What's the OA ?
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06 May 2010, 12:59
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The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

Given: $$2<m<p$$ and $$\frac{p}{m}\neq{integer}$$. $$p=xm+r$$. q: $$r=?$$

(1) The greatest common factor of m and p is 2 --> $$p$$ and $$m$$ are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as $$\frac{p}{m}\neq{integer}$$), then the remainder must be more than 1. Sufficient.

(2) The least common multiple of m and p is 30 --> if $$m=5$$ and $$p=6$$, then remainder=1=1 and thus the answer to the question will be NO. BUT if $$m=10$$ and $$p=15$$, then remainder=5>1 and thus the answer to the question will be YES. Two different answers. Not sufficient.

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06 May 2010, 13:04
Thanks for the explanation, Bunuel !

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06 May 2010, 22:16
Bunuel wrote:
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

Given: $$2<m<p$$ and $$\frac{p}{m}\neq{integer}$$. $$p=xm+r$$. q: $$r=?$$

(1) the greatest common factor of m and p is 2 --> p and m are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since remainder is not zero (as $$\frac{p}{m}\neq{integer}$$), then remainder must be more than 1. Sufficient.

(2) the least common multiple of m and p is 30 --> if m=5 and p=6, remainder=1=1, answer to the question would be NO. BUT if m=10 and p=15 remainder=5>1 answer to the question would be YES. Two different answers. Not sufficient.

Perfect explanation! But slightly tough question to be approached within 2 minutes.

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17 Jul 2013, 05:01
Bunuel wrote:
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

Given: $$2<m<p$$ and $$\frac{p}{m}\neq{integer}$$. $$p=xm+r$$. q: $$r=?$$

(1) The greatest common factor of m and p is 2 --> $$p$$ and $$m$$ are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as $$\frac{p}{m}\neq{integer}$$), then the remainder must be more than 1. Sufficient.

(2) The least common multiple of m and p is 30 --> if $$m=5$$ and $$p=6$$, then remainder=1=1 and thus the answer to the question will be NO. BUT if $$m=10$$ and $$p=15$$, then remainder=5>1 and thus the answer to the question will be YES. Two different answers. Not sufficient.

Bunuel, one question: Why is the answer A? If p is 12 and m is 10, then the GCF is 2 and the reminder would be 1. What am I missing? Many thanks!

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17 Jul 2013, 05:10
BankerRUS wrote:
Bunuel wrote:
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

Given: $$2<m<p$$ and $$\frac{p}{m}\neq{integer}$$. $$p=xm+r$$. q: $$r=?$$

(1) The greatest common factor of m and p is 2 --> $$p$$ and $$m$$ are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as $$\frac{p}{m}\neq{integer}$$), then the remainder must be more than 1. Sufficient.

(2) The least common multiple of m and p is 30 --> if $$m=5$$ and $$p=6$$, then remainder=1=1 and thus the answer to the question will be NO. BUT if $$m=10$$ and $$p=15$$, then remainder=5>1 and thus the answer to the question will be YES. Two different answers. Not sufficient.

Bunuel, one question: Why is the answer A? If p is 12 and m is 10, then the GCF is 2 and the reminder would be 1. What am I missing? Many thanks!

12 divided by 10 gives the remainder of 2, not 1: 12=1*10+2.

Hope it helps.
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01 Nov 2013, 10:52
Bunuel wrote:
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

Given: $$2<m<p$$ and $$\frac{p}{m}\neq{integer}$$. $$p=xm+r$$. q: $$r=?$$

(1) The greatest common factor of m and p is 2 --> $$p$$ and $$m$$ are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as $$\frac{p}{m}\neq{integer}$$), then the remainder must be more than 1. Sufficient.

(2) The least common multiple of m and p is 30 --> if $$m=5$$ and $$p=6$$, then remainder=1=1 and thus the answer to the question will be NO. BUT if $$m=10$$ and $$p=15$$, then remainder=5>1 and thus the answer to the question will be YES. Two different answers. Not sufficient.

I just have one doubt which seems going against my basic concepts when 15 is divided by 10, it has same result as when 3 is divided by 2, and when you divide 15 by 10, you do not take it as 3 divided by 2. The remainders of both the fractions should 1, but you are taking it otherwise.

Pl explain this part.

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02 Nov 2013, 04:24
ankit41 wrote:
Bunuel wrote:
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

Given: $$2<m<p$$ and $$\frac{p}{m}\neq{integer}$$. $$p=xm+r$$. q: $$r=?$$

(1) The greatest common factor of m and p is 2 --> $$p$$ and $$m$$ are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as $$\frac{p}{m}\neq{integer}$$), then the remainder must be more than 1. Sufficient.

(2) The least common multiple of m and p is 30 --> if $$m=5$$ and $$p=6$$, then remainder=1=1 and thus the answer to the question will be NO. BUT if $$m=10$$ and $$p=15$$, then remainder=5>1 and thus the answer to the question will be YES. Two different answers. Not sufficient.

I just have one doubt which seems going against my basic concepts when 15 is divided by 10, it has same result as when 3 is divided by 2, and when you divide 15 by 10, you do not take it as 3 divided by 2. The remainders of both the fractions should 1, but you are taking it otherwise.

Pl explain this part.

Not sure I understand what you mean. Can you please elaborate? Thank you.

Both 15/10 and 3/2 is 1.5 but the remainder when 15 is divided by 10 is 5 and the remainders when 3 is divided by 2 is 1.
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02 Nov 2013, 11:16
Bunuel wrote:
ankit41 wrote:
Bunuel wrote:
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

Given: $$2<m<p$$ and $$\frac{p}{m}\neq{integer}$$. $$p=xm+r$$. q: $$r=?$$

(1) The greatest common factor of m and p is 2 --> $$p$$ and $$m$$ are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as $$\frac{p}{m}\neq{integer}$$), then the remainder must be more than 1. Sufficient.

(2) The least common multiple of m and p is 30 --> if $$m=5$$ and $$p=6$$, then remainder=1=1 and thus the answer to the question will be NO. BUT if $$m=10$$ and $$p=15$$, then remainder=5>1 and thus the answer to the question will be YES. Two different answers. Not sufficient.

I just have one doubt which seems going against my basic concepts when 15 is divided by 10, it has same result as when 3 is divided by 2, and when you divide 15 by 10, you do not take it as 3 divided by 2. The remainders of both the fractions should 1, but you are taking it otherwise.

Pl explain this part.

Not sure I understand what you mean. Can you please elaborate? Thank you.

Both 15/10 and 3/2 is 1.5 but the remainder when 15 is divided by 10 is 5 and the remainders when 3 is divided by 2 is 1.

Why is the remainder of 15/10 is 5 and not 1? we know that 15/10 is same as 3/2. What I am doing while calculating remainders, I simplify the fraction if it can be further simplified like in the case of 15/10, I simplified the fraction to 3/2 and then I will find the remainder, it comes out to be 1 but If I donot simplify the fraction the remainder comes out to be 5.

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02 Nov 2013, 21:08
ankit41 wrote:
Why is the remainder of 15/10 is 5 and not 1? we know that 15/10 is same as 3/2. What I am doing while calculating remainders, I simplify the fraction if it can be further simplified like in the case of 15/10, I simplified the fraction to 3/2 and then I will find the remainder, it comes out to be 1 but If I donot simplify the fraction the remainder comes out to be 5.

Hmm. Well what you are doing is not entirely correct. When you simplify the given fractions, keep in hand the common factor which factored out from both the Numerator and the Denominator. For eg, in $$\frac{15}{10}$$ we factored out 5 from both the places. Thus, the remainder of the simplified fraction, as you correctly said is 1. Just multiply this back with the common factor, i.e. in this case 5. And hence, the final remainder is 5.

You cannot just cancel out factors and expect the same remainder.

For eg, if your logic were true, we would have Remainder of $$\frac{16}{6}$$ = Remainder of $$\frac{8}{3}$$ which is definitely not the case.
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Re: The integers m and p are such that 2<m<p, and m is not a [#permalink]

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14 Jan 2018, 06:26
Bunuel wrote:
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

Given: $$2<m<p$$ and $$\frac{p}{m}\neq{integer}$$. $$p=xm+r$$. q: $$r=?$$

(1) The greatest common factor of m and p is 2 --> $$p$$ and $$m$$ are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since the remainder is not zero (as $$\frac{p}{m}\neq{integer}$$), then the remainder must be more than 1. Sufficient.

(2) The least common multiple of m and p is 30 --> if $$m=5$$ and $$p=6$$, then remainder=1=1 and thus the answer to the question will be NO. BUT if $$m=10$$ and $$p=15$$, then remainder=5>1 and thus the answer to the question will be YES. Two different answers. Not sufficient.

look at condition 1
even can be divided by even
6/2=3
result is odd.
but because m and p are different by a number time of 2. r can be 2.

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Re: The integers m and p are such that 2<m<p, and m is not a   [#permalink] 14 Jan 2018, 06:26
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