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08 Jan 2022, 12:50
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Difficulty:
95%
(hard)
Question Stats:
38% (02:27) correct
62%
(02:38)
wrong
based on 89
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If a and b are positive integers greater than 1 and least common multiple of 5a and 5b is 60, what is the value of a?
(1) Greatest Common Divisor of 3a and 3b is 6.
(2) Twice of Greatest Common Divisor of 3a and 3b is 3b.
(1) Greatest Common Divisor of 3a and 3b is 6.
(2) Twice of Greatest Common Divisor of 3a and 3b is 3b.
Updated on: 09 Jan 2022, 08:20
Originally posted by ReedArnoldMPREP on 08 Jan 2022, 14:13.
Last edited by ReedArnoldMPREP on 09 Jan 2022, 08:20, edited 1 time in total.
Last edited by ReedArnoldMPREP on 09 Jan 2022, 08:20, edited 1 time in total.
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Quote:
Okay, so I have divisibility here. The givens are that a,b are POS INT, and both > 1.
60 is the LCM of 5a and 5b... That means 5a and 5b are both factors of 60.
Least common multiple of 5a and 5b is 60. A little tricky... Let's think. 60 = 2^2 * 3 * 5 ... the 5 overlap is confusing me, let me just write the factors of 60:
1*60
2*30
3*20
4*15
5*12
6*10
5a and 5b must both be multiples of 5. So 5a and 5b are either 60, 30, 20, 15, or 10 (can't be '5' since a,b≠1).. Meaning a and b are either 12, 6, 4, 3, or 2.
But I know not every pair works (5a = 20 and 5b = 10, for instance, doesn't work since 20 is the LCM of 20 and 10, not 60).
Okay, I probably COULD keep going through possible cases here... I glance at the statements and they kind of bum me out so I am going to keep working through possible cases:
a and b could be: 12 and 6, 12 and 4, 12 and 3, or 12 and 2; 6 and 4, 4 and 3.
{NOTE: One of the easiest things to mix up on a problem like this is applying a constraint of '5a and 5b' to 'a and b.' a and b's least common multiple can be something other than 60. 5a and 5b are the ones constrainted with a LCM of 60, so I got these possible a and b pairs by thinking about 5a and 5b being: 60 and 30, 60 and 20, 60 and 15, 60 and 12, and 60 and 10; 30 and 20; and 20 and 15....Technically, I suppose '60 and 60' so a and b both 12 would also have to be considered).
Quote:
I had a momentary thought about this "Well I won't know which is a," ... but I'm not sure a and b won't be the *same* (...though it seems very unlikely since the only situation I can think that works would be a=b=12). Still, I'm cautious.
Multiplying my pairs of possible a and b and seeing which have a GCD of 6 tells me a and b could be: 12 and 2, or 6 and 4. But I do not know which is which.
Eliminate A and D
Quote:
Had to pause for a moment to understand what this meant. It tells me the GCD of 3a and 3b is *half* of 3b. But there's no divisor or a number between half that number in itself... So 3a *must* be that number [NOTE: see correction below]. That is 3a = (1/2)3b. Which means
a = (1/2)b
...Okay so I stopped here. Something wasn't clicking. The statements CANNOT contradict themselves and neither case that was possible in statement 1 was possible in statement 2. 12 and 6 are the only statements that work for statement 2, but 12 and 6 don't work for statement 1...
I think this question is poorly designed. Someone correct me if I'm missing something!
CORRECTION!
I made a bad assumption. It is true that one situation that follows statement 2's rule is when b = 2a. But that is not the *only* scenario that follows b's rule. It works whenever b has some prime factors of a *and then one extra two*.
So b=12 and a = 6 is a case that works, **but so does a = 6 and b = 4.**
Here's another fun part. In either case a = 6.
Great question! Tricky stuff.
Updated on: 09 Jan 2022, 09:13
Originally posted by gmatophobia on 09 Jan 2022, 02:06.
Last edited by gmatophobia on 09 Jan 2022, 09:13, edited 1 time in total.
Last edited by gmatophobia on 09 Jan 2022, 09:13, edited 1 time in total.
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Official Explanation
Step 1: Analyse Question Stem
* a and b are positive integers greater than 1
* LCM(5a, 5b) = 60
* Thus, we can infer that LCM(a, b) = \(\frac{16}{5}\) = 12 = \(2^2 * 3^1\)
* Since a and b are > 1
* Some of the possible combinations of a and b are (4, 6), (3, 4), (2, 12), (6,12), and (12, 12)
We need to find the value of a.
Statement 1: GCD of 3a and 3b =6
Gcd(3a, 3b) =6
* Thus, the GCD(a, b) =\(\frac{6}{3}\) = 2
* We know that LCM of (a, b) = \(2^2 * 3^1\)
Thus, we can infer that possible combinations of (a, b) can be (4, 6), (6, 4), (2, 12) or (12, 2)
So, there can be multiple values of a.
Hence, statement 1 is not sufficient, we can eliminate Options A and D.
Statement 2: Twice of GCD of 3a and 3b is 3b.
2*GCD(3a, 3b) = 3b
GCD(a, b) = \(\frac{b}{2}\)
Let’s see which of the combination satisfy this condition.
download.png [ 6.32 KiB | Viewed 3864 times ]
The only cases that satisfy the condition are (6, 4) (6, 12)
And in both these case the value of a = 6.
Hence, statement 2 is sufficient, the correct answer is Option B.
Step 1: Analyse Question Stem
* a and b are positive integers greater than 1
* LCM(5a, 5b) = 60
* Thus, we can infer that LCM(a, b) = \(\frac{16}{5}\) = 12 = \(2^2 * 3^1\)
* Since a and b are > 1
* Some of the possible combinations of a and b are (4, 6), (3, 4), (2, 12), (6,12), and (12, 12)
We need to find the value of a.
Statement 1: GCD of 3a and 3b =6
Gcd(3a, 3b) =6
* Thus, the GCD(a, b) =\(\frac{6}{3}\) = 2
* We know that LCM of (a, b) = \(2^2 * 3^1\)
Thus, we can infer that possible combinations of (a, b) can be (4, 6), (6, 4), (2, 12) or (12, 2)
So, there can be multiple values of a.
Hence, statement 1 is not sufficient, we can eliminate Options A and D.
Statement 2: Twice of GCD of 3a and 3b is 3b.
2*GCD(3a, 3b) = 3b
GCD(a, b) = \(\frac{b}{2}\)
Let’s see which of the combination satisfy this condition.
Attachment:
download.png [ 6.32 KiB | Viewed 3864 times ]
The only cases that satisfy the condition are (6, 4) (6, 12)
And in both these case the value of a = 6.
Hence, statement 2 is sufficient, the correct answer is Option B.
