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I don't know what the given solution is from the Total GMAT Math Book but what I did was look for an answer that made the numerator zero. For any other integer for \(m\) (given \(n=1\)) we can likely find a value for \(p\) that satisfies the equation. The only situation where this doesn't happen is if the numerator is zero. It's impossible to divide zero by any number and get \(2\). Hope this helps!

Guess I wasn't fast enough on my reply lol.
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akhilright
Thanks BB, That was pretty simple, I just complicated the math in a away where I got lost! Thanks

BB: members/member-3.html
Bunuel: members/member-73391.html

Two different people. :-D
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akhilright
Thanks BB, That was pretty simple, I just complicated the math in a away where I got lost! Thanks

BB: members/member-3.html
Bunuel: members/member-73391.html

Two different people. :-D
OOOpps my BAD....! apologies...!
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This is what I did initially and would warn ppl not to complicate unnecessarily:

I took the fraction \(\frac{m-n}{p}=2\) and simplified it to \(m-n=2p\) further as given n=1, I simplified it and got \(m=2p+1\) then I looked on the answers and trying to find an even number as m seems to be an odd one!...and so I created a mental block and cudn think of simply looking for a number which could make the numerator 0 !!!!!!

I end up doing such things and get a Q44! on the real test

Phew!
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akhilright
This is what I did initially and would warn ppl not to complicate unnecessarily:

I took the fraction \(\frac{m-n}{p}=2\) and simplified it to \(m-n=2p\) further as given n=1, I simplified it and got \(m=2p+1\) then I looked on the answers and trying to find an even number as m seems to be an odd one!...and so I created a mental block and cudn think of simply looking for a number which could make the numerator 0 !!!!!!

I end up doing such things and get a Q44! on the real test

Phew!

Notice that 2p+1 (m) would be and odd number if p is an integer.
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akhilright
This is what I did initially and would warn ppl not to complicate unnecessarily:

I took the fraction \(\frac{m-n}{p}=2\) and simplified it to \(m-n=2p\) further as given n=1, I simplified it and got \(m=2p+1\) then I looked on the answers and trying to find an even number as m seems to be an odd one!...and so I created a mental block and cudn think of simply looking for a number which could make the numerator 0 !!!!!!

I end up doing such things and get a Q44! on the real test

Phew!

Notice that 2p+1 (m) would be and odd number if p is an integer.

OK, Finally I get the reason why we should not use 'p' in our calculation because we don't know if its an integer or not! 8-) Thanks BUNUEL!
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This is what I did initially and would warn ppl not to complicate unnecessarily:

I took the fraction \(\frac{m-n}{p}=2\) and simplified it to \(m-n=2p\) further as given n=1, I simplified it and got \(m=2p+1\) then I looked on the answers and trying to find an even number as m seems to be an odd one!...and so I created a mental block and cudn think of simply looking for a number which could make the numerator 0 !!!!!!

I end up doing such things and get a Q44! on the real test

Phew!

Notice that 2p+1 (m) would be and odd number if p is an integer.

OK, Finally I get the reason why we should not use 'p' in our calculation because we don't know if its an integer or not! 8-) Thanks BUNUEL!

You can also see that the only time the equation will not work is if p=0.
So make p=0 and solve for m:

\(\frac{m-n}{p}=2\)

\(\frac{m-1}{0}=2\)

m - 1 = 0
m = 1
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I'm surprised the question stem does not say p != 0. Typically, most of the GMAT questions I have seen specify such condition.
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I'm surprised the question stem does not say p != 0. Typically, most of the GMAT questions I have seen specify such condition.

The fact that \(p\neq{0}\) is implied in the condition \(\frac{m-n}{p}=2\). If \(p\) were \(0\) that expression would not be defined.

Hope it's clear
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as others have already clarified , I will try to put it this way.

For all the other options except 1 LHS of the equation in the question stem is "some number" (since we don't know the value of P and whether it is an integer or not is also an unknown".

For the valuei choice D i.e. 1, Irrespective of the value of P ' It has to be 0 since the numerator becomes zero and so It can not be the value of M since RHS is not 0
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lchen

You can also see that the only time the equation will not work is if p=0.
So make p=0 and solve for m:

\(\frac{m-n}{p}=2\)

\(\frac{m-1}{0}=2\)

m - 1 = 0
m = 1

That's not correct: (m-n)/p=2 implies that p does not equal to zero, division by 0 is not allowed.
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Its simple.. if the value of m=1, the LHS becomes 0 and so cannot be 2,

no m cannot be 1
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lchen

You can also see that the only time the equation will not work is if p=0.
So make p=0 and solve for m:

\(\frac{m-n}{p}=2\)

\(\frac{m-1}{0}=2\)

m - 1 = 0
m = 1

That's not correct: (m-n)/p=2 implies that p does not equal to zero, division by 0 is not allowed.
That is exactly why Ichen chose to solve the equation for m when p=0, because we are looking for the value of m that is not possible
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HBSdetermined
This is what I did initially and would warn ppl not to complicate unnecessarily:

I took the fraction \(\frac{m-n}{p}=2\) and simplified it to \(m-n=2p\) further as given n=1, I simplified it and got \(m=2p+1\) then I looked on the answers and trying to find an even number as m seems to be an odd one!...and so I created a mental block and cudn think of simply looking for a number which could make the numerator 0 !!!!!!

I end up doing such things and get a Q44! on the real test

Phew!

I did the exact same thing and substituted x with 0.5 to get a different answer. So in such cases are we suppose to assume that x is an integer?
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Deconstructing the Question

We are given

\(\frac{m-n}{p}=2\)

and

\(n=1\)

We need the value of m that is not possible.

The fastest check is to test the choice that makes the numerator equal to zero.

Step-by-step

Substitute \(n=1\):

\(\frac{m-1}{p}=2\)

Now test \(m=1\).

Then the left side becomes

\(\frac{1-1}{p}=\frac{0}{p}=0\)

But the equation says the value must be \(2\).

So this is impossible.

Therefore \(m=1\) is not a possible value.

Answer: D
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