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Difficulty:
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Question Stats:
65% (01:55) correct
35%
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wrong
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If a, b, and c are integers and a*b^2/c is a positive even integer, which of the following must be true?
I. ab is even
II. ab > 0
III. c is even
A. I only
B. II only
C. I and II
D. I and III
E. I, II, and III
I. ab is even
II. ab > 0
III. c is even
A. I only
B. II only
C. I and II
D. I and III
E. I, II, and III
Orange08
Given: \(\frac{a*b^2}{c}=even>0\) --> \(ab^2=c*even=even\) --> either \(a\) is even or \(b\) or both.
I. \(ab\) is even --> according to the above this must be true;
II. \(ab>0\) --> not necessarily true, \(b\) could be positive as well as negative (for example \(a=1\), \(c=1\) and \(b=-2\));
III. \(c\) is even --> not necessarily true, see above example.
Answer: A (I only).
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General Discussion
Posts: 9
Countdown
ab2/c is positive which means a and C >0 or a and c<0 since b2 is positive
Case 1 - a=3 b=2 c=3 or c=6 where ab2 /2 would be even when c= 3 or c=6
lets see III option where C is even as per our assumption C can be even and odd, hence it is out.
lets see option II ab>0 a and b can positive or negative hence sign of ab cannot be determined hence it is out
So options B,C,D,E are gone and ans is A
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Countdown
I believe the expression is \(ab^2/c\) is a positive integer.
Now we know,
1. even * even = even (2*4 = 8)
2. even * odd = even (2*3 = 6)
3. even / even = even or odd (8/2 = 4 , 6/2 = 3)
4. even / odd = even (6/3 = 2)
5. odd * odd = odd
6. odd / odd = odd (if the are divisible)
7. odd / even = <not divisible> as the denominator will always have an extra 2.
Coming back to my question, my result is even. Hence my numerator must be even, as odd numerator can never give even result. (we are considering divisible integers only)
\(ab^2\) is even, hence either a is even or \(b^2\) is even. If \(b^2\) is even then b is even. Hence ab = even. (I) is true.
now \(\sqrt{b^2}\) = +/- b, so ab can be > 0 or < 0, (II) is false
From rule 3 and 4, c can be even or odd. so (III) is false.
Answer A.
Countdown
Given: a, b, and c are integers and ab^2/c is a positive even integer.
Evaluate each option:
I. ab is even.
ab^2/c to be positive even integer ab^2 must be even, from which it follows that either a or b must be even, thus ab has to be even.
II. ab > 0
Since b is squared in our expression, then we can say nothing about its sign, thus we cannot say whether ab is positive.
III. c is even
The easiest one: ab^2/c = integer/c = even --> c could be even as well as odd.
Answer: A.
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Orange08
We are given that (a*b^2)/c is a positive even integer. Therefore, a*b^2 must be even. (If a*b^2 is odd, (a*b^2)/c can’t ever be even.)
Now recall that the product of an even number and any integer is even, so either a or b, or both, must be even. Thus we see that ab must be an even integer. However, ab DOES NOT have to be greater than zero, since a could be -2 and b could be 1. Finally, we see that c does not have to be even, since a could be -2, b could be 1, and c = -1. Thus, only Roman numeral I must be true.
Answer: A
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Can be solved less than 1 min conceptually!
We know that (these formulas can come handy to save time):
Case 1: even/even = even or odd
Case 2: even/odd = even
I. ab is even: in both cases, numerator needs to be EVEN, so this is a MUST be TRUE condition
II. ab>0: Not necessarily, to produce positive result, we need either both positive or both negative as numerator and denominator, so this does not need to be true always
II. c is even: From Case 1 and Case 2, it can be seen that the denominator can be even or odd. So, c can be either even or odd, so not true always!
Answer A. I only!
We know that (these formulas can come handy to save time):
Case 1: even/even = even or odd
Case 2: even/odd = even
I. ab is even: in both cases, numerator needs to be EVEN, so this is a MUST be TRUE condition
II. ab>0: Not necessarily, to produce positive result, we need either both positive or both negative as numerator and denominator, so this does not need to be true always
II. c is even: From Case 1 and Case 2, it can be seen that the denominator can be even or odd. So, c can be either even or odd, so not true always!
Answer A. I only!
Orange08
Given: \(a*b^2/c\) is an even integer
It means the numerator ab^2 is even i.e. either a or b or both must have 2 as a factor. Hence ab certainly has 2 as a factor and is even.
Hence, I must be true.
Given: \(a*b^2/c\) is positive
b^2 will be positive in any case - whether b is positive or negative.
This means a and c are either both positive or both negative.
So ab may be positive or negative since a can be positive or negative and b can be positive or negative. Hence ab doesn't need to be positive.
c needn't be even. c must be a factor of ab^2 but it may or may not have 2 as a factor.
e.g. c = 3, a = 6, b = 1
Here ab^2/c will be even integer though c is odd.
Answer (A)
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Last visit: 29 Jun 2026
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Bunuel
Bunuel, I have a question on II. Why are we not considering the scenario of ab=0. That's when I thought II is "must be true". The question says nothing about a,b,c being != 0. 0 is an integer as well. Could you please explain? Thank you!
KarishmaB
Responding to a pm:
Engineer1
Statement 2 says ab > 0
What that means is that when we multiply a and b, we get a positive number. Is it possible that even one of a and b is 0?
The moment one of them becomes 0, ab becomes 0 and cannot be positive.
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Engineer1
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KarishmaB
KarishmaB, thanks for the quick response.
Yes, for this exact reason ab>0 is a must be true statement for ab (or ab^2) to be a positive integer, is what I was referring to. Therefore the answer should be I & II (C).
Engineer1
Think about it: Is it necessary that ab > 0 ? You are right that ab cannot be 0 but can't we have ab < 0?
Say a = 4, b = -1 and c = 2 (all integers while ab < 0)
Then
\(\frac{a*b^2}{c} = \frac{4 * (-1)^2}{2} = 2\) (positive even integer)
So ab < 0 is also possible. It is NOT NECESSARY that ab > 0
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Cases:
All numbers can be negative
b can be negative (a & c positive)
a & c can be -ve
Even/Odd = Even and Even/Even= Even
ab^2/c= Even
ab^2= C*Even = Even
ab^2=Even
Therefore, ab has to be even
ab can be less than zero since (from above) b can be negative and a can be positive - Hence ab>0 is incorrect
c can be even or odd since,
even/odd = even and
even/even = even
since ab^2 is even c can either be even or odd. Therefore 'c is even' is incorrect
IMO A
Please correct logic wherever required. Thanks
All numbers can be negative
b can be negative (a & c positive)
a & c can be -ve
Even/Odd = Even and Even/Even= Even
ab^2/c= Even
ab^2= C*Even = Even
ab^2=Even
Therefore, ab has to be even
ab can be less than zero since (from above) b can be negative and a can be positive - Hence ab>0 is incorrect
c can be even or odd since,
even/odd = even and
even/even = even
since ab^2 is even c can either be even or odd. Therefore 'c is even' is incorrect
IMO A
Please correct logic wherever required. Thanks
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