## the integers m and p are such that 2<m<p

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### the integers m and p are such that 2<m<p

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
RonPurewal
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### Re: q3

whew! this one is a serious cacophony of rephrasing and interpretation.

let's translate:
m is not a factor of p.

if m were a factor of p, then the remainder upon dividing p by m would be 0.

therefore, we can translate the above statement as follows:
"the remainder upon dividing p by m is not 0."
in other words, it's an integer greater than 0.

--

the question:
is r > 1 ?

here's a HIGH-LEVEL INTERPRETATION of this problem.

if the remainder WERE 1, then p would be 1 more than a multiple of m.
if this is the case, then p and m CANNOT have any common factors, other than 1. (this is so because all factors of m are factors of (p-1), which is a multiple of m; a number greater than 1 can't be a factor of both (p-1) and p, which are consecutive integers.)

therefore, if m and p have common factors, then the answer to this question is YES.

(note that the converse is not necessarily true: even if there are no common factors, the answer still could be yes. for instance, 17 divided by 6 leaves a remainder of 5, even though 17 and 6 have no common factors. but, if we can establish that there are common factors, then that's enough to show that the answer is Yes.)

--

statement (1)

if this is true, then m and p have the factor 2 in common, so, YES.
sufficient.

--

statement (2)

this doesn't tell you whether m and p have common factors.
if m = 5 and p = 6, for instance, then r = 1.
if m = 10 and p = 15, then r = 5, which is > 1.
insufficient.

ans (a)

--
RonPurewal
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### Re: q3

secondary solution: (because the primary solution above is fairly obnoxious)

JUST PLUG IN NUMBERS.

statement (1)

let's just PICK A WHOLE BUNCH OF NUMBERS WHOSE GCF IS 2 and watch what happens. let's try to make the numbers diverse.
say,
4 and 6
6 and 8
8 and 10
10 and 12
...
4 and 10
6 and 14
6 and 16
8 and 18
8 and 22
...
in all nine of these examples, the remainders are greater than 1. in fact, there is an obvious pattern, which is that they're all even, since the numbers in question must be even.

in fact, i just thought of this, which is a much nicer, more ground-level approach to statement one:
in statement 1, both m and p are even. therefore, the remainder is even, so it's greater than 1.

done.

sufficient.

--

statement (2)
just pick various numbers whose lcm is 30.
notice the numbers selected above:
5 and 6 --> remainder = 1
10 and 15 --> remainder = 5 > 1
insufficient.
mba012012
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### Re: the integers m and p are such that 2<m<p

Hello,

in fact, there is an obvious pattern, which is that they're all even, since the numbers in question must be even.

No where in the question mentioned that M and P are even. It just shows that M>2 and P>M.

Thanks !
mba012012
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### Re: the integers m and p are such that 2<m<p

Sorry .

I understood now. Because GCF is 2, both should be even.
RonPurewal
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### Re: the integers m and p are such that 2<m<p

mba012012 wrote:Sorry .

I understood now. Because GCF is 2, both should be even.

so we're all good now, then.
rachelhong2012
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### Re: the integers m and p are such that 2<m<p

This problem reminds me of this problem:

in-x-and-y-are-positive-integers-such-that-x-8y-12-t902.html

Stacey mentioned an interesting principle/pattern in that post:

If one number is b units away from another number, and b is a factor of both numbers, the GCF of the two numbers is b

this principle jumped out to my mind when I saw the first statement

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

I was thinking to myself:

hmm, can the converse of the principle be true?
that if the GCF of the two numbers is b (in this case it's 2), and b (2) is a factor of both numbers (m&p), then m is 2 units away from p? or that the remainder is 2 when p/m?

Of course, this isn't true since Ron's picking numbers that are not 2 units away from each other prove I'm wrong.
RonPurewal
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### Re: the integers m and p are such that 2<m<p

rachel, i like the way you analyzed that situation a lot (even though you wound up coming to the conclusion that your tentative "rule" doesn't actually work).

that's the best way to investigate this kind of stuff -- actually TRY things and see whether they work! it's very refreshing to see a student who is not just trying to memorize a bunch of random facts. keep it up.
krishnan.anju1987
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### Re: the integers m and p are such that 2<m<p

I can't believe it. I understood why b was insufficient immediately but kept going at a even after reading the explanations.. for quite some time and now I finally understand hopefully.

If p=10 and m=6. p/m =10/6 and remainder is 4. I kept dividing both by their common factor and getting 5/3, remainder =2,

sometimes as in the case where p=8 and m=6, p/m =8/6=4/3
the remainder is 1

and thus I thought this was insufficient.

Please tell me the mistake I made was to divide the numerator and denominator by the common factor.
RonPurewal
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### Re: the integers m and p are such that 2<m<p

krishnan.anju1987 wrote:I can't believe it. I understood why b was insufficient immediately but kept going at a even after reading the explanations.. for quite some time and now I finally understand hopefully.

If p=10 and m=6. p/m =10/6 and remainder is 4. I kept dividing both by their common factor and getting 5/3, remainder =2,

sometimes as in the case where p=8 and m=6, p/m =8/6=4/3
the remainder is 1

and thus I thought this was insufficient.

Please tell me the mistake I made was to divide the numerator and denominator by the common factor.

ya, that's not how remainders work. a remainder represents a concrete number of things left over, so you can't "reduce" the numbers you're dividing.

e.g.
let's say that i have 26 hot dogs, and i'm putting them into packs of 8.
in this case, the remainder when 26 is divided by 8 is 2.
what does this mean?
it means that, when i put 26 hot dogs into packs of 8, i have 2 hot dogs left over (after making three full packs of 8).
in this situation, i think it's clear why we can't "reduce" the 26/8 to 13/4 ... because, well, that doesn't reflect the actual situation anymore. we aren't trying to parcel out thirteen of any item into groups of four; we're trying to parcel out twenty-six items into groups of eight.
gkashyap
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### Re: the integers m and p are such that 2<m<p

The question is solved and I have no doubt about the explanation.
I had a good laugh reading Ron's last explanation of hot dog packaging. When I Imagine it, I can't stop laughing! Ron, you've got a great sense of humor!
RonPurewal
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### Re: the integers m and p are such that 2<m<p

gkashyap wrote:The question is solved and I have no doubt about the explanation.
I had a good laugh reading Ron's last explanation of hot dog packaging. When I Imagine it, I can't stop laughing! Ron, you've got a great sense of humor!

Thanks, but I usually have to think about things like hot dogs in order to understand remainders.
NinaP494
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### Re: the integers m and p are such that 2<m<p

Tim's proof in that thread helped me realize that since m is not a factor of p, their gcm should be less than their difference. Thus they should be at least 2 places apart and hence r>2. Hope my logic is correct.
RonPurewal
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### Re: the integers m and p are such that 2<m<p

"GCM"?
no such animal.

there's a LCM and a GCF; presumably you meant one of those.

i suppose you could define a "LCF", but that would just be = 1 for any pair of integers.

"GCM" is not a thing that exists, because there is no limit on how big multiples can be. "GCM" would be ∞.