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Re: If a and b are positive integers such that a < b, is b even? [#permalink]
17 Dec 2012, 23:33

Expert's post

PraPon wrote:

If a and b are positive integers such that a < b, is b even?

(1) \frac{b}{2}-\frac{a}{2} is an integer.

(2) \frac{3b}{4}-\frac{a}{2} is an integer.

Statement 1 leads us to two possibilities: i) either the two i.e.a/2 and b/2 are of the form x.5 AND y.5 respectively ii) both of them to be integers.

Both of these possibilities lead us to the integer form. So B CAN be even and CANNOT be even.

Statement 2- We have to keep in mind that a and b are integers. Hence 3b/4 can only be an integer or of the form x.75 or x.25 Since (xx.75 or xx.25)-yy.5 CANNOT be an integer, therefore B has to be an integer such that 3b/4 is an integer. For that to happen, B has to be a multiple of 4. The only possibility when 3b/4 falls in the form of xx.5 is when B is an even integer.

OR

a/2 can only be either entirely an integer or of the form yy.5. So in such cases, 3b/4 - a/2 will be an integer ONLY when b is an even integer.

Re: If a and b are positive integers such that a < b, is b even? [#permalink]
18 Dec 2012, 01:24

2

This post received KUDOS

Expert's post

If a and b are positive integers such that a < b, is b even?

(1) \frac{b}{2}-\frac{a}{2} is an integer --> \frac{b}{2}-\frac{a}{2}=integer --> b-a=2*integer=even. From b-a=even it follows that either both a and b are even or both odd. Not sufficient.

(2) \frac{3b}{4}-\frac{a}{2} is an integer --> \frac{3b}{4}-\frac{a}{2}=integer --> 3b-2a=4*integer=even. Since 2a=even, then we have that 3b-even=even --> 3b=even --> b=even. Sufficient.

Re: If a and b are positive integers such that a < b, is b even? [#permalink]
18 Dec 2012, 08:39

Nothing can be simpler than Bunuel's explanation!! Thanks Bunuel!!

I pretty much didn't like Manhattan GMAT's explanation. I guess it was too lengthy and convoluted.

Manhattan GMAT explanation:- For this yes/no question, or goal is to try to find a definitive answer: either b is always even or b is always something other than even (odd or a fraction / decimal). If b is even only some of the time, then that information would be insufficient to answer the question.

This is also a theory question; on such questions, we can try numbers or we can use theory. We can also test some numbers initially in order to help ourselves figure out or understand the theory more thoroughly and then use theory to help guide us through the rest of the problem.

(1) INSUFFICIENT: We can test cases here to get started. First, let’s test the case where both a and b are even. If b = 4 and a = 2, then b/2-a/2=2-1=1. This makes sense using theory: we know that dividing an even integer by 2 will result in another integer. The variables a and b are both integers, so dividing each one by two will also yield integers, and one integer minus another integer will yield a third integer. Using both real numbers and theory, we have proved that the result will be an integer, so it’s possible for b to be even.

Could b also be odd? Dividing an odd number by 2 yields some integer followed by the decimal 0.5 (for example 3/2 = 1.5). If we subtract one x.5 number from another, we’ll still get an integer. For instance, if b = 5 and a = 3, then b/2-a/2 = 2.5 – 1.5 = 1. It’s also possible, then, for b to be odd. Since b can be either even or odd, this statement is not sufficient.

We have also now picked up something useful about the theory: an integer minus an integer will yield another integer. A non-integer minus another non-integer with the same decimal value (e.g., 2.5 – 1.5) will also yield an integer.

(2) SUFFICIENT: We’re going to test even and odd cases here again. We already determined during statement 1 that a/2 will be an integer if a is even. What would need to be true in order for 3b/4 to be an integer as well? The value of b would have to be some multiple of 4 (in order to “cancel out” the 4 on the bottom of the fraction). We can try the same numbers we tried last time: b = 4 and a = 2. In this case, 3b/4-a/2 = 3 – 1 = 2. It’s possible, then, for b to be even.

Can b be odd? There are two possible cases to test: odd b and odd a, or odd b and even a. An even value for a will result in an integer for a/2; for this to make statement 2 true, we would need 3b/4 to be an integer as well. 3b/4 will never result in an integer when b is odd, however, because an odd divided by an even will never be an integer. For example, if b = 5 and a = 2, then 3b/4-a/2 = 15/4 – 1 = not an integer. We can dismiss the case where a is even and b is odd.

What about the case where both a and b are odd? If a is odd, then a/2 will be some number ending in 0.5. Can we make 3b/4 also end in 0.5, so that we’ll get an integer when subtracting the two? Let's try some odd positive integer possibilities for b: 3b/4 could equal 3/4, 9/4, 15/4, and so on, or the decimal equivalents 0.75, 2.25, 3.75, and so on. The pattern here alternates between 0.75 and 0.25; we cannot get 0.5. We can’t, then, get an integer value for 3b/4-a/2 as long as b is odd.

Re: If a and b are positive integers such that a < b, is b even? [#permalink]
18 Dec 2012, 09:35

Expert's post

Bunuel wrote:

If a and b are positive integers such that a < b, is b even?

(1) \frac{b}{2}-\frac{a}{2} is an integer --> \frac{b}{2}-\frac{a}{2}=integer --> b-a=2*integer=even. From b-a=even it follows that either both a and b are even or both odd. Not sufficient.

(2) \frac{3b}{4}-\frac{a}{2} is an integer --> \frac{3b}{4}-\frac{a}{2}=integer --> 3b-2a=4*integer=even. Since 2a=even, then we have that 3b-even=even --> 3b=even --> b=even. Sufficient.

Re: If a and b are positive integers such that a < b, is b even? [#permalink]
22 Mar 2013, 14:34

freespiritfox wrote:

If a and b are positive integers such that a < b, is b even?

(1) B/2- A/2 is an integer.

(2) 3*B/4 - A/2 is an integer

(1) B/2- A/2 is an integer. B/2- A/2=i B- A=2i 2i is even and can be obtained as Even-Even or Odd-Odd so 1 is not sufficient

(2) 3*B/4 - A/2 is an integer 3*B/4 - A/2=i 3*B - 2A= 4i 4i is even and can be obtained as Even-Even or Odd-Odd. Now consider that the second term is even (2A) so the other must be even also. So, 3*B is even; can we say that B is also even? The answer is yes, because Even = Odd*Even = 3(odd)*B(even)

So 2 is sufficient

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