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If integer C is randomly selected from 20 to 99, inclusive

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If integer C is randomly selected from 20 to 99, inclusive

by RBBmba@2014 » Wed Feb 11, 2015 10:45 pm
If integer C is randomly selected from 20 to 99, inclusive. What is the probability that c^3-c is divisible by 12?
A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3

OA:C

P.S: Experts(Mitch/Rich/Matt/Dave and others) - could you please share your analysis & explanation? Much thanks in advance!
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by DavidG@VeritasPrep » Thu Feb 12, 2015 1:53 am
First, note we have 80 numbers in this range. The denominator will be have to be a factor of 80, so get rid of anything that isn't. B and E are out.

In order to be divisible by 12, a number needs to have 2^2 * 3. So we need two 2's and a 3.

Now let's break down c^3 - c. Rewrite as c(c^2 - 1) = c(c + 1)(c - 1). Well, these are three consecutive numbers. So we want to know the probability that three consecutive numbers will contain 2^2 * 3 among them. Any three consecutive numbers will always contain a multiple of 3, so all we need are the two 2's.

There are two ways this can happen. First, if c is odd, then c- 1 and c + 1 will both be even. Two evens, by definition must contain at least two 2's between them. So anytime c is odd, we'll have our two 2's. Number of odds from 20 - 99? We've got 80 numbers and half of them are odd, so that's 40.

The second way we can get our two 2's is if c itself is a multiple of 4. We've got 80 numbers and 1/4 of them will be multiples of 4, so that's another 20.

So we've got 40 + 20 = 60 desired outcomes.

60/80 = 3/4. Answer is C
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by DavidG@VeritasPrep » Thu Feb 12, 2015 1:57 am
This thread contains a similar question, and you might find the discussion helpful:

https://www.beatthegmat.com/probability- ... tml#739216
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by GMATGuruNY » Thu Feb 12, 2015 4:08 am
RBBmba@2014 wrote:If integer C is randomly selected from 20 to 99, inclusive. What is the probability that c^3-c is divisible by 12?
A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3
c³ - c = c(c² - 1) = c(c+1)(c-1) = (c-1)(c)(c+1).
In other words, c³ - c is the product of 3 CONSECUTIVE INTEGERS.

We need to determine the probability that the product of 3 consecutive integers will be a multiple of 12.
Strategy:
WRITE OUT consecutive options for (c-1)(c)(c+1) and LOOK FOR A PATTERN.

c=20 --> 19*20*21
c=21 --> 20*21*22
c=22 --> 21*22*23
c=23 --> 22*23*24

c=24 --> 23*24*25
c=25 --> 24*25*26
c=26 --> 25*26*27
c=27 --> 26*27*28

All of the products in red are divisible by both 3 and 4 and thus are multiples of 12.
Of every 4 products, 3 are divisible by 12.
Since there are 80 integers between 20 and 99, inclusive, there are 80 options for c and thus 80 options for (c-1)(c)(c+1).
Since 3/4 of these products will be divisible by 12, the probability that c³ - c will yield a multiple of 12 = 3/4.

The correct answer is C.
Last edited by GMATGuruNY on Thu Feb 12, 2015 10:00 am, edited 1 time in total.
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by Brent@GMATPrepNow » Thu Feb 12, 2015 9:38 am
RBBmba@2014 wrote:If integer C is randomly selected from 20 to 99, inclusive. What is the probability that c^3-c is divisible by 12?
A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3
I want to point out that David is using an important property that can be summarized as follows:

The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.

Here are a few more questions where this rule can be applied:
- https://www.beatthegmat.com/if-n-is-an-i ... 76329.html
- https://www.beatthegmat.com/pls-explain-t270792.html
- https://www.beatthegmat.com/divisibility-t276068.html

Cheers,
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by [email protected] » Thu Feb 12, 2015 9:03 pm
Hi RBBmba@2014,

The other posts all explain the "math" behind this prompt, so I won't rehash any of that here. It's important to remember that the GMAT is a test of many skills, including "pattern recognition" and "thoroughness." In many questions, you'll be asked about subjects that you already know, but you'll be asked to deal with the subject in a way that you might not be used to thinking about it.

If you were asked to multiply C(C+1)(C-1), then I'm sure that you could do it. Here though, you have to recognize that same pattern (about C^3 - C) but in reverse. In many cases, "weird looking" information, equations, etc. can be rewritten and you're EXPECTED to rewrite them. If you find yourself thinking "that math will take forever", then there is most assuredly some other way to get to the solution. Be ready to "play" with information, take notes and look for patterns. In many cases, the big "breakthrough" that you need to solve a given question will come right from those activities.

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by RBBmba@2014 » Thu Feb 12, 2015 10:52 pm
Experts - Much appreciate your detail reply and explanation! Really helpful.

Few quick clarifications for Mitch and Dave -

@Dave - I think, product of 3 consecutive positive integers will always be divisible by 6(=3x2). Right ? So, how we can solve this question in terms of this concept ?

P.S: I replied in the thread you shared, it'd be great if you can check that out as well!

@Mitch - As going by the pattern, I believe, we're to consider this one also -

c=99 --> 98*99*100. And it's a valid combination in this question and divisible by 12.Right ?
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by GMATGuruNY » Fri Feb 13, 2015 3:49 am
RBBmba@2014 wrote: @Mitch - As going by the pattern, I believe, we're to consider this one also -

c=99 --> 98*99*100. And it's a valid combination in this question and divisible by 12.Right ?
Correct.
The greatest four options for (c-1)(c)(c+1) are as follows:
c=96 --> 95*96*97
c=97 --> 96*97*98
c=98 --> 97*98*99
c=99 --> 98*99*100
Notice that the pattern repeats:
The three products in red are divisible by both 3 and 4 and thus are multiples of 12.
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by DavidG@VeritasPrep » Fri Feb 13, 2015 7:42 am
I think, product of 3 consecutive positive integers will always be divisible by 6(=3x2). Right ?

Exactly right. Any 3 consecutive numbers will have at least one multiple of 2 and have exactly one multiple of 3, and thus, when multiplied, will be a multiple of 6. So because we know that we're guaranteed to have a 2 and a 3, the only thing we really need to determine in this question is whether we have a second 2, as 12 = 2*2*3. Anytime you have three consecutive numbers that go EVEN ODD EVEN, we know we'll have our second 2, because the evens will both contain a 2. So one desired scenario is when the middle number, c, is ODD.

If, however, we have ODD EVEN ODD, then in order to have two 2's, that EVEN number will have to be a multiple of 4, because clearly, the ODDs won't contribute any 2's. So the other desirable scenario is when c is a multiple of 4.
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by RBBmba@2014 » Fri Feb 13, 2015 10:35 am
Mitch/Dave - Much appreciate your replies! :-)
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by Matt@VeritasPrep » Sun Feb 15, 2015 5:31 pm
RBBmba@2014 wrote: product of 3 consecutive positive integers will always be divisible by 6(=3x2). Right ? So, how we can solve this question in terms of this concept ?
Another way of thinking about this:

* Every SECOND number is divisible by 2
* Every THIRD number is divisible by 3

So if you have three consecutive integers, at least one of them divides by 2 and at least one of them divides by 3, so their product must divide by 6.

Here's a neat follow up question that builds on this concept:
Is p divisible by 24?

S1:: p is prime
S2:: p is a two-digit integer
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by RBBmba@2014 » Sun Feb 15, 2015 9:37 pm
Matt@VeritasPrep wrote:
Is p divisible by 24?

S1:: p is prime
S2:: p is a two-digit integer
Matt - just curious that how we can relate this question to this discussion! Could you please elaborate ?

IMO OA is A and answer is a definite NO. Correct me please if wrong!
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by GMATinsight » Mon Feb 16, 2015 3:27 am
RBBmba@2014 wrote:If integer C is randomly selected from 20 to 99, inclusive. What is the probability that c^3-c is divisible by 12?
A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 1/3

OA:C
c^3-c is one of the favorite expressions that GMAT has been asking questions from therefore we must understand that such questions can be done in the following manner

1) Substituting the values and finding a pattern among the acceptable/rejectable values of c

2) Understand the standard properties of consecutive numbers
e.g.
- Product of any three consecutive integers will always be divisible by 6 because every set of three consecutive integers include atleast one Even Integer and one Integer multiple of 3 thereby making the product as divisible by 3
- Product of every 3 consecutive integers WITH TWO EVEN AND ONE ODD Integers will always be divisible by 24 because two consecutive even Integers will have one number a multiple of 2 and other a multiple of 4 (at-least) and three consecutive numbers will include one multiple of 3 as well

Such observation in long run make the learning more effective therefore such inferences are very important in order to handle such questions more effectively.

I hope it contributes among various methods of solution of the same question

All the best!!!
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