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
Solution:
"For a product of integers to be even, at least one of those integers needs to be even. So the question is asking: is either one of m, p, or t even ? "
That is exactly what we look for.
(1)
t - p = p - m
t = 2p - m
-don't know if p is even or odd, but 2p is even.
-don't know if m is even or odd
t = 2p - even = even
t = 2p - odd = odd
N/S
(2)
t - m = 16
t = 16 +m
-don't know if m is even or odd
t = 16 + odd = odd
t = 16 = even = even
N/S
(1) and (2)
t = 2p - m
t = 16 + m
2p - m = 16 + m
16 = 2(p - m)
8 = p - m
-don't know whether p or m are even or odd.
N/S
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3 comments:
in this question we get p=(t+m)/2 now for p to be an integer p+m shud be an integer and hence either both shud be even or both should be odd
in first case since both wud be even the answer wud be 1 is sufficient
in second case wen t & m both wud be odd AND diffrnt their sum will be even and p=(some even no.)/2 which is again even and hence p will be even and answer will again be 1 is sufficient.
- vineet
Hi Vineet,
Let's say you are true up to your point and now here we will try to justify your point with an example.
So consider this example:-
m = 3, p = 5 and t = 7 [All three are odd intergers]
Now according to statement (1),
p = (m + t)/2
Does our stated values support the statement 1?
YES it is. So product of 3 odd integers will always be an odd integer.
Okay what if all three integer values are EVEN?
Lets say m = 2, p = 4 and t = 6
Again it will satisfy statement (1).
Product of 3 EVEN INTEGERS is an EVEN INTEGER.
So, it means statement (1), leads us to both YES and NO.
Hence Statement (1) is insufficient.
I am sure now you can prove this for statement (2) and for both statement on your own.
Let me know if more help is required on this topic.
Good Luck....
- BTG760
duh... how cud i even write that.. my bad!
guess got confused between even no/2 wud be an integer rather than always even!!
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