If A and B are integers, is the product AB even?

Question

I. A + B = √(10A+3B)
II. (66B-2)/A= 13A-B

chetan2u · Answer

If A and B are integers, is the product AB even?

If any of A and B is even, AB will be even.

I. A + B = √10A+3B
A-2B=√10A.... Square both sides..  A^2+4B^2-4AB=10A...........A^2=10A+4AB-4B^2=2(5A+2AB-2B^2) ..
Therefore, A is a multiple of 2. So, AB is even.
Sufficient
Edited version.. A+B=√(10A+3B)
Square both sides..  A^2+B^2+2AB=10A+3B ... Let us see if both A and B are odd
 odd^2+odd^2+2*odd*odd=10*odd+3*odd....O+O+E=E+O.....E+E=E+O...E=O ..Not possible
So at least one is even. Which one is not required, we can say AB will be even
Sufficient

II. (66B-2)/A= 13A-B
66B-2=13A^2-AB...... AB=66B-2-13A^2 ..
Therefore if A is even  even=even*B-2-13*even .. possible
If A and B are odd.. odd*odd=66*odd-even-13*odd......odd=even-even-odd ... possible
So AB can be even or Odd
Insufficient

A

kiran120680 · Answer

Edited Statment I.

Manat · Answer

chetan2u  why is B not sufficient?

(66B-2)/A= 13A-B

66* Any number will be even

So (Even-Even)/A = 13A-B, since A & B are integers means A can divide 66B-2, So A is even

Moving to RHS

13*A - B =Even
13*EVEN-B=EVEN
Even-B=Even

Hence B is also Even.

chetan2u · Answer

(66B-2)/A is an integer does not mean A is even..
Say B is 2, then (66*2-2)/A = 130/A is integer.. so A can be 13,2,5 ... A can be odd or even.
Even if B is odd or even , A can be 1 and odd..

kiran120680 · Answer

In the Statment I, Square root is for (10A+3B) not for only 10A. Please consider the statement and explain.

mangamma · Answer

Hope, you took the statment I wrongly, I have seen the same question eGMAT. square root is for (10A+3B), not only for 10A

chetan2u · Answer

The answer for edited question...

If A and B are integers, is the product AB even?

If any of A and B is even, AB will be even.

I. A+B=√(10A+3B)
Square both sides..  A^2+B^2+2AB=10A+3B ... Let us see if both A and B are odd
 odd^2+odd^2+2*odd*odd=10*odd+3*odd....O+O+E=E+O.....E+E=E+O...E=O ..Not possible
  So at least one is even. Which one is not required, we can say AB will be even  
Sufficient

II.  \frac{(66B-2)}{A}= 13A-B 
66B-2=13A^2-AB...... AB=66B-2-13A^2 ..
Therefore if A is even  even=even*B-2-13*even .. possible
If A and B are odd.. odd*odd=66*odd-even-13*odd......odd=even-even-odd ... possible
So AB can be even or Odd
Insufficient

A

Saupayan · Answer

For statement 2:
The issue with this approach is it doesn't take into account the RHS constraint of A and B. There are only a certain set of values that will satisfy this equation. We need to test if withing that constrained set, we can have A and B as even/odd whatever. That is time consuming. I doubt this is a good GMAT representative problem

If A and B are integers, is the product AB even?

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