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Even/Odd & Divisibility Problem

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by Brian@VeritasPrep » Thu Sep 23, 2010 8:57 am
Hello, everyone:

So David@VeritasPrep and I have a little informal competition going regarding who is the more-thanked poster on the site here, and I've seen him posting some original questions on a few of these threads to some success. I figure it's about time I did a few of my own, so let's start with this one! I'll be back with a solution later today if you need it.



If a, b and c are integers, is ab/c an odd integer?


(1) When a is divided by c, the quotient is an odd integer.


(2) When b is divided by c, the quotient is an odd integer.
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Source: — Data Sufficiency |
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by selango » Thu Sep 23, 2010 9:14 am
stmt1,

When a is divided by c, the quotient is an odd integer.

No info about b.

If b is even then ab/c is even

If b is odd then ab/c is odd

Insuff

stmt2,

When b is divided by c, the quotient is an odd integer.

No info about a.

If a is even then ab/c is even

If a is odd then ab/c is odd

Insuff

Combining 1 and 2,

There are 2 scenarios in this case.

1. a=10,b=14,c=2

ab/c is even

2. a=21,b=9,c=3

ab/c is odd

Insuff

Pick E
Last edited by selango on Thu Sep 23, 2010 9:18 am, edited 1 time in total.
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by this_time_i_will » Thu Sep 23, 2010 9:16 am
should be E.
It is easy to see why I and II, independently are not sufficient.

For I & II:
6*10/2 = even
9*15/3 = odd
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by clock60 » Thu Sep 23, 2010 11:33 am
i wonder is it possible to solve some formal way
(1) a=c(2k+1) where k is integer
(2) b=c(2m+1) where m is integer
a*b/c=c*(2k+1)c(2m+1)/c=c*(2k+1)(2m+1) and now it depend on the value of c, if c is odd then all phrase is odd, but in the problem the value of c is not fixed so E is the answr
guys do you think this is right approach?
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by Brian@VeritasPrep » Thu Sep 23, 2010 2:58 pm
Great answers, everyone!

Clock60, you can certainly solve this one formally (and algebraically, in a way). Here's how I'd do it:

Statement 1 says that:

a/c = an odd integer

So...if you multiply both sides by c, you get:

a = c * odd

That means that, if c is even, so is a. But if c is odd, then a is odd. This statement is not sufficient.


Statement 2 says that:

b/c = an odd integer

b = c * odd

Similarly, this tells us that b and c are either both even or both odd, but we don't know which.


Taken together, we know that a = c * odd and that b = c * odd, so:

ab/c = [c(odd) * c(odd)] / c

We can divide out one of the cs in the numerator with the c in the denominator, but we still have c * odd * odd, meaning that if c is even, the whole number will be even, but that if c is odd the whole number will be odd. Therefore, we can algebraically demonstrate that neither statement is sufficient, even if taken together. The answer is E.
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by HPengineer » Thu Sep 23, 2010 3:32 pm
opps i thought an even/even always produces an odd result?? 10/2 = 5 and 6/2 = 3
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by Brian@VeritasPrep » Thu Sep 23, 2010 4:54 pm
Hey, HP:

But try 8/2 = 4. There aren't hard-and-fast even/odd rules for division, so you'll need to try those out for yourself.

Basically, even just means "divisible by 2". But to divide a number and still have a factor of 2, you have to have more evens in the numerator than in the denominator:

8/2 = 2*2*2/2 You'll cancel one two out of the numerator and still have some remaining

6/2 = 2*3/2 You'll cancel the only two out of the numerator and be left with only an odd remaining
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by HPengineer » Thu Sep 23, 2010 5:51 pm
Got it will be more diligent in my picking of values... Good problem thanks!
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