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Even Odd Problem

by melguy » Sat Oct 22, 2016 7:20 pm
Hello all

Please help me with an even odd problem.

If a,b,c are integers and ab + c is odd. Which of the following must be true?

1. a+c is odd
2. b+c is odd
3. abc is even

My challenge here is not the problem itself but the timing. I have figured out the problem by drawing a table and and testing with brute force but it took me good 4 - 5 minutes. Is there a smarter way to figure out the problem in under 2 minutes.

Appreciate your help!
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by [email protected] » Sun Oct 23, 2016 9:21 am
Hi melguy,

You're essentially responsible to check 4 possible situations, so it's not clear how you spent 4-5 minutes on this prompt. Maybe you TESTED difficult values or perhaps you need to work on your organization.. You might find it helpful to use the built-in Number Properties to your advantage though.

To start, we know that A, B and C are all INTEGERS and (A)(B) + C = ODD.

That equation is the sum of two values - to end in an ODD sum, one of the 'parts' must be EVEN and the other must be ODD. Thus, we could have...

(A)(B) = even and C = odd
(A)(B) = odd and C = even

Breaking this down further, in the first situation...
1st option A = even, B = even, C = odd
2nd option A = even, B = odd, C = odd
3rd option A = odd, B = even, C = odd

In the second situation...
4th option A = odd, B = odd, C = odd
However, the prompt tells us that (A)(B) + C = ODD, so this 4th option is NOT allowed.

We're asked which of the Roman Numerals are true. Since you have not listed the 5 answer choices, we're forced to check all 3.

1. A+C is odd

Not in the 3rd option.

2. B+C is odd

Not in the 2nd option.

3. ABC is even

In all 3 of the permutable options, ABC IS even, so Roman Numeral 3 is always true.

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Last edited by [email protected] on Sun Oct 23, 2016 5:01 pm, edited 1 time in total.
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by fiza gupta » Sun Oct 23, 2016 10:19 am
ab + c = odd => even(e) + odd(o) = odd
one has to be even and one odd from ab and c

(i) ab = odd , c = even
ab to be odd both has to be odd
so we get a = odd , b = odd and c = even - (i)

(ii) ab = even , c =odd
ab = even => either even or both
a = even b = even c odd -(ii)
b = even a = odd c = odd - (iii)
a = even b = odd c = odd -(iv)

a+c is odd
(iii) conditions will make it false

2. b+c is odd
first three conditions will make it false

3. abc is even
always true in all above options
or we can do it directly
because ab + c = odd
either has to be even and product will be always even.
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by fiza gupta » Sun Oct 23, 2016 10:27 am
To start, we know that A, B and C are all INTEGERS and (A)(B) + C = ODD.

That equation is the sum of two values - to end in an ODD sum, one of the 'parts' must be EVEN and the other must be ODD. Thus, we could have...

(A)(B) = even and C = odd
(A)(B) = odd and C = even

Breaking this down further, in the first situation...
1st option A = even, B = even, C = odd
2nd option A = even, B = odd, C = odd
3rd option A = odd, B = even, C = odd

In the second situation...
4th option A = odd, B = odd, C = odd

We're asked which of the Roman Numerals are true. Since you have not listed the 5 answer choices, we're forced to check all 3.

1. A+C is odd

Not in the 3rd or 4th options.

2. B+C is odd

Not in the 2nd or 4th option.

3. ABC is even

Not in the 4th option

Thus, NONE of the Roman Numerals is always true.
in 4th option : a = odd , b odd and c= odd
ab + c = will be even because odd*odd = odd
odd +odd = even
Fiza Gupta
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by [email protected] » Sun Oct 23, 2016 5:02 pm
Hi fiza gupta,

Good eye. I've updated my solution.

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by vineet.nitd » Tue Oct 25, 2016 1:14 am
What is the implication of ab + c being odd?


It implies that one of the two components (ab OR c) is even, while the other is odd.

If that is true their product is also going to be even, since one component is even for sure.
Hence, 3 must be true.

Statements 1 and 2 have symmetry as "a" AND "b" could behave in the same way given the original expression. So, we know that either statements 1 and 2 must both be true OR must both be false.


Now let's test statement 1, given ab + c is odd :
If a is even, b is odd, c is odd => Statement 1 is true
If a is odd, b is even, c is odd=> Statement 1 is not true

So, both statements 1 and 2 may or may not be true.
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by Brent@GMATPrepNow » Tue Oct 25, 2016 8:00 am
melguy wrote: If a,b,c are integers and ab + c is odd. Which of the following must be true?

1. a+c is odd
2. b+c is odd
3. abc is even
You should be able to test all 8 possible values within 2 minutes. Here we go:
A. a = EVEN, b = EVEN, c = EVEN: In this case ab + c is EVEN
B. a = ODD, b = EVEN, c = EVEN: In this case ab + c is EVEN
C. a = EVEN, b = ODD, c = EVEN: In this case ab + c is EVEN
D. a = EVEN, b = EVEN, c = ODD: In this case ab + c is ODD
E. a = ODD, b = ODD, c = EVEN: In this case ab + c is ODD
F. a = ODD, b = EVEN, c = ODD: In this case ab + c is ODD
G. a = ODD, b = ODD, c = EVEN: In this case ab + c is ODD
H. a = ODD, b = ODD, c = ODD: In this case ab + c is EVEN

Since we're told that ab + c is ODD, we can eliminate cases A, B, C and H
Now, let's test the remaining cases.

1. a+c is odd
D. a = EVEN, b = EVEN, c = ODD: In this case a + c is ODD
E. a = ODD, b = ODD, c = EVEN: In this case a + c is ODD
F. a = ODD, b = EVEN, c = ODD: In this case a + c is EVEN
G. a = ODD, b = ODD, c = EVEN: In this case a + c is ODD
As we can see, a+c NEED NOT be ODD
So, statement 1 is not true

2. b+c is odd
D. a = EVEN, b = EVEN, c = ODD: In this case b + c is ODD
E. a = ODD, b = ODD, c = EVEN: In this case b + c is ODD
F. a = ODD, b = EVEN, c = ODD: In this case b + c is ODD
G. a = ODD, b = ODD, c = EVEN: In this case b + c is ODD
Great, b+c is always odd.
Statement 2 is true

3. abc is even
D. a = EVEN, b = EVEN, c = ODD: In this case abc is EVEN
E. a = ODD, b = ODD, c = EVEN: In this case abc is EVEN
F. a = ODD, b = EVEN, c = ODD: In this case abc is EVEN
G. a = ODD, b = ODD, c = EVEN: In this case abc is EVEN
Great, abc is always even
Statement 3 is true

RELATED VIDEO
- Testing Possible Cases: https://www.gmatprepnow.com/module/gmat ... /video/839
Brent Hanneson - Creator of GMATPrepNow.com
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by Wakenour » Wed Oct 26, 2016 5:59 am
Brent@GMATPrepNow wrote:
melguy wrote: If a,b,c are integers and ab + c is odd. Which of the following must be true?

1. a+c is odd
2. b+c is odd
3. abc is even
You should be able to test all 8 possible values within 2 minutes. Here we go:
A. a = EVEN, b = EVEN, c = EVEN: In this case ab + c is EVEN
B. a = ODD, b = EVEN, c = EVEN: In this case ab + c is EVEN
C. a = EVEN, b = ODD, c = EVEN: In this case ab + c is EVEN
D. a = EVEN, b = EVEN, c = ODD: In this case ab + c is ODD
E. a = ODD, b = ODD, c = EVEN: In this case ab + c is ODD
F. a = ODD, b = EVEN, c = ODD: In this case ab + c is ODD
G. a = ODD, b = ODD, c = EVEN: In this case ab + c is ODD
H. a = ODD, b = ODD, c = ODD: In this case ab + c is EVEN

Since we're told that ab + c is ODD, we can eliminate cases A, B, C and H
Now, let's test the remaining cases.

1. a+c is odd
D. a = EVEN, b = EVEN, c = ODD: In this case a + c is ODD
E. a = ODD, b = ODD, c = EVEN: In this case a + c is ODD
F. a = ODD, b = EVEN, c = ODD: In this case a + c is EVEN
G. a = ODD, b = ODD, c = EVEN: In this case a + c is ODD
As we can see, a+c NEED NOT be ODD
So, statement 1 is not true

2. b+c is odd
D. a = EVEN, b = EVEN, c = ODD: In this case b + c is ODD
E. a = ODD, b = ODD, c = EVEN: In this case b + c is ODD
F. a = ODD, b = EVEN, c = ODD: In this case b + c is ODD
G. a = ODD, b = ODD, c = EVEN: In this case b + c is ODD
Great, b+c is always odd.
Statement 2 is true

3. abc is even
D. a = EVEN, b = EVEN, c = ODD: In this case abc is EVEN
E. a = ODD, b = ODD, c = EVEN: In this case abc is EVEN
F. a = ODD, b = EVEN, c = ODD: In this case abc is EVEN
G. a = ODD, b = ODD, c = EVEN: In this case abc is EVEN
Great, abc is always even
Statement 3 is true

RELATED VIDEO
- Testing Possible Cases: https://www.gmatprepnow.com/module/gmat ... /video/839
eagle eye... and I guess I say that because I literally have problems with my eyes :D (tired of Diamox already but what I can do?)
Last edited by Wakenour on Tue Feb 07, 2017 10:06 am, edited 3 times in total.
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by melguy » Thu Oct 27, 2016 4:36 am
Thanks everyone for your valuable input!

I second that! Brent you have got eagle eyes :)
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by Matt@VeritasPrep » Thu Oct 27, 2016 11:58 pm
I think we're missing an easy approach here.

There's nothing special about a or b: we could just as easily replace one with the other. (For example, ab + c could be 1*2 + 3 or 2*1 + 3.)

From that, we can deduce that either 1 and 2 are BOTH true or BOTH false. Picking a simple set of numbers like my sets above (a = 1, b = 2, c = 3 and a = 2, b = 1, c = 3) lets us this out: they're both false!

From there, we've got III. abc will be even if ANY of a, b, and c are even. Let's see what happens if none of them is even:

a = 1, b = 3, c = 5

But this gives us ab + c = 1*3 + 5 = 8, which isn't odd. (As you can guess, this generalizes to any set of odd numbers: odd*odd + odd = even.)

Since we CAN'T have a, b, c all odd, we must have at least one even, or a * b * c = even.

So III only is true, and we're done!
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