When Logic Trumps Formulas: Even/Odd Integers

by on April 23rd, 2013

Some of the most difficult quant questions on the GMAT are the ones that use the fewest formulas.

One such category of problems involve even/odd integer arithmetic. Although they involve few calculations, they can require a surprising amount of logical reasoning. Take this problem, for instance:

If a and b are integers, and m is an even integer, is ab/4 an integer?

(1) a + b is even.
(2) m/(ab) is an odd integer.

Before reading our solution, go ahead and attempt to solve this problem on your own.

The key to tackling this problem lies in using a few basic facts that you probably already know about even/odd integers. Here is a quick refresher on the rules:

  1. An even number can only be formed by the sum of either 2 odd numbers (odd + odd = even), or 2 even numbers (even + even = even).
  2. An odd number can only be formed by the sum of an odd and even number (odd + even = odd, or even + odd = odd).
  3. An even number can only be formed by multiplication in three ways: even·odd, odd·even, and even·even.
  4. An odd number can only be formed by multiplication in one way: odd·odd = odd.

The challenge is to use these four basic statements to deduce some hidden information.

Assuming that statement 1 alone is true, we know that a + b must be even, so according to rule #1, a and b must either both be even, or both be odd. Let’s consider the two possible scenarios:

Suppose a and b were both odd. We could plug in a = 3 and b = 1 to get ab/4 = 3·1/4 = 3/4; this is clearly not an integer. However, if both a and b were even, we could plug in a = 24 and b = 2, so that ab/4 = 24·2 / 4 = 12, which is in fact an integer. Because different plug-ins lead to different answers, we cannot answer with a definite yes or a definite no. Statement 1 alone is insufficient, so eliminate choices A and D.

Looking at statement 2 alone, we can assume that m/ab = odd. If we multiply both sides of by ab, we get m = ab·odd. Now, we’re told that m is an even integer, which means that even = ab·odd. According to rule #3, it is impossible to get an even number by multiplying two odd numbers; at least one number must be even. Therefore, since m is even, we must conclude that ab is also even.

If ab is even, does that give us any clue as to whether ab/4 must be an integer? If we try plugging in one set of numbers, such as m = 24, a = 4, and b = 2, then ab = 8, and ab/4 = 8/4 = 2. The answer for this set of plug ins is Yes, and so we might be tempted to conclude that the answer will always be Yes.

There is no such general rule, however, that states that an even number (ab) divided by another even number (4) will always be an integer. As an even integer, ab will be divisible by 2 without a remainder, but there is nothing that implies that it must be divisible by 4. For example, suppose a = 2, b = 1, and m = 6. In this case, ab = 2, and ab/4 = 2 / 4 = 1/2, which would result in a No answer to the question stem. Sometimes statement 2 gives us Yes, and other times No, so statement 2 alone is insufficient; we can proceed to eliminate choice B.

This changes when we look at both statements combined. We know from statement 1 that both a and b must either both be even or both odd, but statement 2 also tells us that ab is even. The only way to reconcile both statements is if both a and b are even; and if both are even, then they both must contain factors of 2. Their product, ab, will therefore contain two factors of 2, and so ab will contain the factor 2·2 = 4, and hence will be divisible by 4. This means that ab/4 must definitely be an integer. Since we have a definite Yes answer, both statements, when combined, are sufficient.

As you can see, problems with no formulas can still prove to be quite tricky. Often with the GMAT, it’s not about what formulas you remember, but about what you can logically deduce.

With that in mind, try out this challenge problem:

If a, b, and c are all integers, is ab+bc+ca+a^2 odd?

(1) a + 2b + c is even.
(2) a·b·c is odd.

(Check back in a few days for the solution in the comments section!)


  • Answer is B.

    In statement 1, a, b & c can be all odd or all even. So, will result into YES when all are odd and NO when all are even. 

    In statement 2, a, b & c all are odd. So, these will result into only YES.

    Hence, B. Am I write??

  • I dunno about which option to choose as I am a complete newbie but the either statements alone will guarantee the question.

    (1) 2b is even for sure. This implies b can be odd or even. Now as 2b is even and a+2b+c is even, it follows thatt a+c is even. This implies both a and c will be both odd or even. This has no correlation to b. Now we come to the question. ab+bc+ca+a^2.

    ca + a^2 will always be even. (if a is odd, c is odd -> ca and a^2 is odd. odd + odd = even). ab+bc will always be even as it can be written as b(a+c) where a+c is always even.

    (2)a.b.c is odd -> all are odd. sum of 4 odd nos=even...

  • Joel is right....Either of the statements are sufficient to answer the question....

  • Statement 1 : a+2b+c is even...here 2b is even whatever the b might be...So "a+c must be EVEN". Using this in the problem. ab+bc+ca+a^2 = b(a+c)+a(c+a) = (c+a)*(a+b) ... Since (c+a) is EVEN the whole expression in the question ab+bc+ca+a^2 is EVEN...SUFFICIENT.
    Statement 2: a.b.c is odd...so all the individual terms are ODD..So using "ODD * ODD = ODD...ODD + ODD = EVEN" we get to know that ab,bc,ca,a^2 are ODD...And Ex: Let a=1,b=3,c=5...3+15+5+1 = 24  which is EVEN...so we have only one aster for this statement so SUFFICIENT...
    ANSWER: D..... Both are sufficient to answer the question.

  • ab+bc+ca+a^2 --> Simplifying this: b(a+c) + a(a+c)

    (1) a + c = (Even Number) - 2b
    2b has to be an even number. Any number multiplied by 2 is even.
    Hence, (a+c) is an Even Number since (Even Number - Even Number = Even Number)
    Substituting this in the simplified eq. above, we get b*(Even Number) + a*(Even Number). This is nothing but Even Number + Even Number which is again EVEN.
    This Statement is Sufficient.

    (2) a.b.c is ODD. The only way this is possible is when all 3 (a, b, c) are ODD. Substituting back in the original eq.
    ODD (ODD + ODD) + ODD (ODD + ODD)
    = ODD.EVEN + ODD.EVEN --> This is because ODD + ODD = EVEN
    = EVEN + EVEN --> This is because ODD.EVEN = EVEN
    = EVEN
    This is Sufficient Too.

    Hence, Either statements are sufficient. Answer (D)

    • only the second statement is sufficient .,,

  • Wonderful explanation about the properties of numbers. Though you have merely restated what most of us, the way it is presented has me thinking about numbers in a whole new way (and answering quite a few questions too)

    • 1) a, b even or a, b odd (Alone not sufficient)
      2) e . o / e = O => ab = e possibilities: a,b=e a=o,b=e a=e,b=o (Alone not sufficient)
      Combining 1) and 2) a,b should be Even as then e * e will give us 2*2=4 in numerator as a factor hence ab/4 will give an integer

  • The question stem can be rephrased as Is (a+b) x (a+c) odd?

    Statement 1 - a+2b+c is even. Since 2b is even, a+c should be even. Therefore (a+b)(a+c) WILL be even. Sufficient!

    Statement 2 - a.b.c is odd. This implies, all three of a,b, and c are odd. So the expression in the question stem would be (Odd + Odd) X (Odd + Odd), or Even x Even, which is Even. Sufficient! 

    Both statements are sufficient by themselves. Answer - D

  • My answer is D; 

    Question is asking if ab+bc+ca+a2 is odd? we can simplify it like (a+b)(a+c) is odd or not? 
    S1 is  sufficient. Bec. 2b+(a+c)=even. since 2b is even, whatever b is even or odd, (a+c) is even. so  (a+b)(a+c) is even... S1 is sufficient.

    S2 is also sufficient. bec. if a.b.c is odd, three of them must be odd. it means a+b and a+c are even, so (a+b)(a+c) must be even.

  • The correct answer is D) -- congrats to all of you who got this problem correct!

    Here's one way to tackle this problem. First, rewrite the expression by doing a little factoring:

    ab + bc + ca + a^2 --> b(a+c) + a(a+c)
    -->(b+a)(a+c) = (a+b)(a+c)

    Now that we have the expression factored, we see it's simply a product of two terms -- a+b and a+c, both of which are integers. In order to get an odd product, both of them must be odd according to rule #4. So we can re-think this DS question in terms of a new question: are both a+b and a+c odd integers?

    Looking at statement 1) alone, we see that a + 2b + c = even. Let's subtract 2b from both sides of the equation to get a + c = even - 2b. Obviously, 2b must be an even integer, so we get a + c = even - even. When we subtract 2 even integers, the result is also even. So statement 1) alone is telling us that a+c is even. As a result, we can say 'No' with certainty to the question stem, and so we have sufficient data. We're down to choices A/D.

    Looking at statement 2) alone, we see that a*b*c is odd. There is only one way for this to occur: all 3 variables must be odd. (If you can't visualize this, test the possibilities out for yourself. See if you can get an odd product if one or more of the variables are even) Since all 3 variables are odd, a+c = odd + odd = even. The result is an even sum, which again leads us to say 'No' with certainty to the question stem. The correct answer is therefore D).

  • can you not assume that A could equal B?

    for example...A = 1 and also B =1. therefore, 2/4 is not an integer.

  • Ray, the stem asks ab/4=int? if a=1 and b=1 then ab=1 not 2. You have to mult not add.

    • haha right.. 1/4 my quesiton is can you assume A = B?

  • haha right. 1/4. tthats not my quesiton. can you assume A = B and they are not separate integers?

    • There's nothing in the stem that says a and b have to be distinct integers, but if you look at stat 2: m/ab=odd int or m=ab*odd int and we know from the stem that m=even and according to rules of of odd/evens an even=odd*even (or even*even). So the hidden constraint is ab has to be even, thus a and b can't both be 1 or both be any other odd for that matter. However, they could both be the same even int (besides zero from stat 2) (according to the rules above).

  • what level question is this? These type of number property questions are my weakest link, but I solved this one relatively well.

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