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Prime numbers and divisibilty

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by CappyAA » Tue Jan 10, 2012 11:59 am
I'm not 100% I understand the question correctly, but the way I read it:

We have a number, t, that has factors: 2*a*c, a*b, and a*b*c. And we want to know what the smallest number of potential integers t could be divisible by. We also know a, b, and c are prime.

To solve this problem, I am going to assume that t is equal to a*b*c. Since a*b*c is the largest of the factors above, we'll assume this is the largest factor. The second assumption I'll make is that one of the three variables is equal to 2. We only know that a, b, and c are different prime numbers. They don't necessarily have to be different than another factor we already know about.

If we make those assumptions, listing the factors is equal to listing all the combinations of a, b, and c:

1, a, b, c, ab, bc, ac, abc

So the answer is (hopefully) equal to 8

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by user123321 » Tue Jan 10, 2012 12:32 pm
bharti.2010 wrote:If a, b, and c are different prime numbers and if a2c, ab, and abc
are all factors of t, what is the smallest number of positive integers
by which t could be divisible?
8
10
12
14
16
i didnt seem to understand the problem i think you wanted to mention factors as a^2*c,ab,bc

is it 12?

since t is having factors a^2*c,ab,bc
t should be atleast a^2*b*c
so no of factors is 3*2*2 = 12

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by Mike@Magoosh » Tue Jan 10, 2012 2:05 pm
So (a^2)c, ab, and abc are all factors of t. The LCM of those three factors is (a^2)bc, so t must be divisible by (a^2)bc.

Now, we need an important fact about factors. To count the total number of factors of a number t, we must
(1) find the prime factorization
(2) find the set of exponents of the prime factors
(3) add one to each exponent
(4) find the product of those numbers

So, assume the prime factorization of t is (a^2)bc --- t may include more factors, but this at least establishes a minimum.

The exponents of prime factors a, b, and c, comprise the set = {2, 1, 1}

Add one to each --> (3, 2, 2}

Find the product of those numbers: 3*2*2 = 12

Therefore, at a minimum, t has 12 factors. Answer choice C.

Does that make sense? Please let me know if you have any questions about this.

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by ronnie1985 » Tue Jan 10, 2012 8:08 pm
It seems that the the question says that t is divisible by each of (a^2)c, ab, and abc
Which means that t must be divisible by (a^2)bc and then the no of factors = 3*2*2 = 12
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by ameya85 » Wed Jan 11, 2012 12:27 am
I solved this by taking first 3 prime numbers.

let, a = 2
b = 3
c = 5

a^2c = 20
ab = 6
abc = 30

LCM is = 60
Factors of 60 =>
60,30,20,15,12,10,6,5,4,3,2,1

So answer must be 12
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by bharti.2010 » Wed Jan 11, 2012 1:41 pm
OA: C
Thanks everybody for your reply. Now I know where I was wrong?
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by chufus » Wed Jan 11, 2012 7:29 pm
Could you edit the question. As is the answer i 8. but if a2 is a^2 then the answer is 12.
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by ArunangsuSahu » Thu Jan 12, 2012 7:02 am
a,b and c are prime

a^2c, ab, abc are factors of t

we can write like the following

t= a*a*b*c which satisfies above factors..


No of positive factors=(p+1)(q+1)(r+1)=3*2*2=12
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by karthikpandian19 » Sat Jun 16, 2012 4:31 am
I could nt understand the last line of your explanation?

What is p, q, r?
ArunangsuSahu wrote:a,b and c are prime

a^2c, ab, abc are factors of t

we can write like the following

t= a*a*b*c which satisfies above factors..


No of positive factors=(p+1)(q+1)(r+1)=3*2*2=12
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by Mike@Magoosh » Mon Jun 18, 2012 10:25 am
karthikpandian19 wrote:I couldn't understand the last line of your explanation? What is p, q, r?
Dear karthikpandian19

I happy to help here. :)

I think what ArunangsuSahu did relatively quickly, in a few lines, will be a little more clear if you look at the step-by-step procedure I spelled out in my post above. In ArunangsuSahu's post, p & q & r are the exponents of the prime factors of t. ArunangsuSahu is clearly very talented in math, but I could see that his work might not be the most comprehensible if you are still trying to make sense of these topics.

I think you may find this blog article help --- it covers all the basics of this topic thoroughly.
https://magoosh.com/gmat/2012/gmat-math-factors/

You may also find this free video helpful:
https://gmat.magoosh.com/lessons/310-gre ... on-divisor

Here's a free practice question on the same concepts:
https://gmat.magoosh.com/questions/865
When you submit your answer to that question, the following page will have a complete video explanation of the question. Each of Magoosh's 800+ GMAT questions has its own video explanation, for accelerated learning.

I hope all that is helpful. Please let me know if have any further questions.

Mike :)
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