If m, p, and t are positive integers.......

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If m, p, and t are positive integers.......

by chendawg » Fri Mar 04, 2011 3:27 pm
If m, p, and t are positive integers and m < p < t, is the product mpt an even integer?

(1) t - p = p - m

(2) t - m = 16

OA after some discussion.

Not too sure what takeaways I can get from this problem.
Source: — Data Sufficiency |

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by Night reader » Fri Mar 04, 2011 3:38 pm
m,p,t>0 t>p>m, find m*t*p/2 is true? Also find whether m/2 or t/2 or p/2 are true?
st(1) t+m=2p, possible cases
--> t, m are odd <> p is odd or even
--> t, m are even <> p is odd or even // Not Sufficient, as all numbers t,m,p could be odd and some numbers could be even
st(2) t=m+even (16), possible cases
--> if t is odd, m must be odd
--> if t is even, m must be even // Not Sufficient because we avail no information about p which could be odd or even
Combined st(1&2), t,m could be odd and p is then even giving p*m*t as even product, yet when p is odd there's an opposite i.e. p*m*t product is odd OR t,m could be even and p must be even then // Obviously Not Sufficient

IOM E
chendawg wrote:If m, p, and t are positive integers and m < p < t, is the product mpt an even integer?

(1) t - p = p - m

(2) t - m = 16

OA after some discussion.

Not too sure what takeaways I can get from this problem.
not clear why we needed m<p<t here? frankly i paid zero attention to this inequality, assuming that we may have the various ranges between t and m to ignore this inequality condition ... // am I correct any, what's OA?
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by pesfunk » Fri Mar 04, 2011 6:33 pm
From the first point: (Insufficient)

t - p = p - m

=> 2p = t + m

So t + m is even. t and m, both can be either odd or even. Example: (3,1) or (4,2)



From the second point: (Insufficient)

t - m = 16.

This is clearly insufficient since 2 odd numbers or 2 even numbers can have the solution as even.



From 1st and 2nd point (Combined): (Insufficient)

Both the numbers can be (20,4) or (21,5)




chendawg wrote:If m, p, and t are positive integers and m < p < t, is the product mpt an even integer?

(1) t - p = p - m

(2) t - m = 16

OA after some discussion.

Not too sure what takeaways I can get from this problem.

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by Anurag@Gurome » Fri Mar 04, 2011 7:38 pm
Solution:
Now the product mpt will be even if any of m, p, or t is even.
Let us verify this.
Consider first (1) alone.
It means t - p = p - m.
Or t + m = 2p.
Let m = 3, p = 4, t = 5. Here m < p < t and t + m = 2p.
Also, mpt is even.
Next let m = 5, p = 7, t = 9. Here m < p < t and t + m = 2p.
But mpt is odd.
So, nothing definite can be said from (1) alone.
Next, consider (2) alone.
t - m = 16.
Let m = 9, p = 11, t = 25. Here m < p < t, t - m = 16.
Also, mpt is odd.
Next, let m = 9, p = 10, t = 25. Here m < p < t. t - m = 16.
But, mpt is even.
So, even (2) alone is not sufficient.
Next combine both statements together and check.
Let m = 1, p =9, t = 17. Here m < p < t, t - m = 16 and t+m = 2p.
Here mpt is odd.
Next let m = 2, p = 10, t = 18. Here m < p <t, t-m = 16 and t+m = 2p.
But mpt is even.
So nothing definite can be said from (1) and (2) together.

The correct answer is (E).
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by chendawg » Sat Mar 05, 2011 8:05 am
OA is E. Source is OG12.

I figure I can just learn to plug in numbers as efficiently as possible from this problem.