Given ‘a’ and ‘b’ are positive integers, is the number b^2+7b+ab even?

Question

(1) (a+ b)(a - b) is a multiple of 2
(2) (a+2)(a-2) is a multiple of 4

andreagonzalez2k · Answer

First we make a table for the original expresion:

a b expresion
even even even
even odd even
odd even even
odd odd odd

(1) (a+b)(a-b)=a^2-b^2 is even
this implies that both a and b are even or odd. If we look at the table we have different results so it is insufficient.
(2) (a+2)(a-2)=a^2-4 is multiple of 4, so it is even
this implies that a must be even. If we look at the table we have that when a is even the expresion is always even, so sufficient.

IMO B

Ayushi0002 · Answer

Can anyone answer this in detail ?

chetan2u · Answer

a and b are integers.

b^2+7b+ab=b(b+7+a)  will be even when 
(i) b is even. 
OR
  b^2+7b+ab=b(b+7)+ab 
Now b(b+7) will be even in every case as b and b+7 will give one as odd and other as even. 
Thus, if a is even, the expression will be even
(ii) a is even.

Thus, if we can show that any of a or b is even, the answer is yes. If both are odd, then answer is no.

Let us see the statements. 
(1) (a+ b)(a - b) is a multiple of 2
It just tells us that both are of same property, that is either both are even or both are odd. 
Insufficient

(2) (a+2)(a-2) is a multiple of 4
If a is odd, then (a+2) and (a-2) will be odd and not even. 
Thus, a is even. 
Sufficient

B

Given ‘a’ and ‘b’ are positive integers, is the number b^2+7b+ab even?

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