Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Does GMAT RC seem like an uphill battle? e-GMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

11 Oct 2014, 13:07

Bunuel : I am finding these type of question difficult to understand .. Can you please simplify a bit more ..explaining how we determine the cases in which the given number is a multiple of the given divisor... please

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

25 Apr 2015, 20:58

shashankp27 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 could see two approaches here and I guess it's good to be consistent with an approach for such type of problems. As proposed by Bunuel on many occasions, I believe in creating a set of N numbers where N is the divisor. In this problem, we have to find out the total numbers divisible by 4 for a function (C-1)C(C+1). Now because the divisor is 4, lets create a set starting with the first integer. In the set-(20,21,22,23), for N = 20,21, and 22, the function (C-1)C(C+1) is divisible by 4. We've 80 numbers b/w 20-99, inclusive and as 80 is divisible by 4 (our divisor), we are safe to assume from above analysis that out 4 number 3 will divide the function. Probability = 3/4.

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

28 Jun 2015, 04:55

shashankp27 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

The answer is C My solution is longer, time consuming. The given expression is (c-1) c (c+1), product of three consecutive integer and it is always divisible by 6. However, we have to find whether this is divisible by 12. we need an additional factor of 2. I used the induction method. When c is 22, 26, 30 and so on up to 98, it is not possible. There are 20 such values. Therefore, the answer 60/80, which is 3/4. If anybody has a crisp solution please....

If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

29 Jul 2015, 17:04

There are 80 numbers from 20-99 inclusive.

c^3 - c = 12k c(c^2-1) = c(c+1)(c-1) = 12k = 2*2*3*k = 4*3*k Make c = 4, then c-1 = 3, and k=5. 4(3)(5)=60 There are 60 numbers that have the given difference divisible by 12 within the given 80 numbers, or 3/4 probability.

If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

28 Mar 2016, 05:50

GuptaDarsh wrote:

Bunuel : I am finding these type of question difficult to understand .. Can you please simplify a bit more ..explaining how we determine the cases in which the given number is a multiple of the given divisor... please

shashankp27 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

1. How many integer between 20 and 99 ? By using formula :\(last = first + (n-1)*1\) => we get: 99 = 20 + n-1 => n = 99-20 +1 = 80 2. \(C^3 - C = (C-1)*C*(C+1)\) => the multiplication of 3 conservative integer => always divisible by 3. Because \(12=4*3\) , our problem is narrowing to a question : "does \(C^3 - C\) is divisible by 4 " 3. It is obvious that :C" could be either even or odd. Let's take a look on these two options. 4. The C is odd 4.1. When C is odd we've \((C-1)*C*(C+1) = (2q)(2q+1)(2q+2) = q(2q+1)(q+1)*4\) 4.2. \(\frac{q(2q+1)(q+1)*4}{4} = q(2q+1)(q+1)\) is integer. Therefore, \(C^3 - C\) => always divisible by 4 if C odd. 4.3. Let's find out how many odds between 21 and 99. By using formula : \(Last = First + (n-1)*2\), we'll find that there are 40 odd. 5. The C is even 5.1. When C is even we've \((C-1)*C*(C+1) = (2q-1)(2q)(2q+1)\)=> by using logic from step 4.2., we can see that, in this, case C isn't divisible by 4 ( ) 5.2. Let's find which even C divisible by 4 between 20 and 98. From "Multiplication Table", we know that \(4*5 = 20, 4*6 = 24, 4*7 = 28 ... 4*24 = 96\). 5.3. How many integers between 5 and 24 ? => \(24 - 5 + 1 = 20\) 6. Now we can calculate our probability : from step 4.3 we got 40 integer and from step 5.3 we got 20 integer. Also we know that between 20 and 99, 80 integer (step 1) 6.1. So\(\frac{20 + 40}{80}\)= 2/3
_________________

I’m not afraid of the man who knows 10,000 kicks and has practiced them once. I am afraid of the man who knows one kick & has practiced it 10,000 times! - Bruce Lee

Please, press the +1 KUDOS button , if you find this post helpful

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

13 Apr 2016, 00:00

Bunuel 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

Two things: 1. There are total of 80 integers from 20 to 99, inclusive: 20, 21, ..., 99. 2. C^3-C=(C-1)*C*(C+1): we have the product of 3 consecutive integers, which is always divisible by 3. So the question basically is whether (C-1)*C*(C+1) is divisible by 4.

Next, the only way the product NOT to be divisible by 4 is C to be even but not a multiple of 4, in this case we would have (C-1)*C*(C+1)=odd*(even not multiple of 4)*odd.

Now, out of first the 4 integers {20, 21, 22, 23} there is only 1 even not multiple of 4: 22. All following groups of 4 will also have only 1 even not multiple of 4 (for example in {24, 25, 26, 27} it's 26, and in {96, 97, 98, 99} it's 98, always 3rd in the group) and as our 80 integers are entirely built with such groups of 4 then the overall probability that C is not divisible by 4 is 1/4. Hence the probability that it is divisible by 4 is 1-1/4=3/4.

Answer: C.

nkhosh wrote:

Would you please explain (1) The formulas you used for finding the # of ODD and "Divisible by 4" outcomes? AND: (2) Why are we looking for the number of ODD integers within 20-99? I'm confused b/c 23 for example is not divisible by 12....

Thank you for the great post

1. There are even # of consecutive integers in our range - 80 (from 20 to 99, inclusive). Out of even number of consecutive integers half is always odd and half is always even, thus numbers of odd integers in the given range is 40.

2. \(# \ of \ multiples \ of \ x \ in \ the \ range =\) \(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

So, for our case as 96 is the last multiple of 4 in the range and 20 is the first multiple of 4 in the range then total # of multiples of 4 in the range is (96-20)/4+1=20.

Or look at it in another way out of 20 consecutive even integers we have half will be even multiples of 4 and half will be even not multiples of 4.

Hope it's clear.

Bunel Cant we solve like the below Their are 7 Multiples of 12 between 20 & 99---> 24,36,48,60,72,84,96. Each multiple has two consecutive integer when substituted in (c-1)*c*(c+1) will be divisible by 12. For ex:- 48 has 47 and 49 as consecutive integers when 47 is substituted--> (47-1)*47*(47+1) is divisible by twelve. same as 49----> (49-1)*49*(49+1) is divisible by 12. Total their are 21 integers as above so probability = 21/80 ~20/80 =1/4.

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

13 Apr 2016, 00:24

akshay4gmat wrote:

Bunel Cant we solve like the below Their are 7 Multiples of 12 between 20 & 99---> 24,36,48,60,72,84,96. Each multiple has two consecutive integer when substituted in (c-1)*c*(c+1) will be divisible by 12. For ex:- 48 has 47 and 49 as consecutive integers when 47 is substituted--> (47-1)*47*(47+1) is divisible by twelve. same as 49----> (49-1)*49*(49+1) is divisible by 12. Total their are 21 integers as above so probability = 21/80 ~20/80 =1/4.

Hi, the method choosen by you will not be answer... This method would give you prob if in three consecutive integers, ONE is a multiple of 12.. But we rae looking at the PRODUCT of these consecutives as div by 12, so it is possible that one number is div by 3 abd other by 4...

lets use your method to find the answer.. the PRODUCT would surely be div by 3, so lets look ONLY for div by 12/3 = 4.. so 20 to 99-- 80/ 4= 20.. now, in a way although not exactly, the same technique that its neighbouring integers can be taken SO total = 20*3=60... so TOTAL 60/80 = 3/4..

if i take 24.. 22,23,24 ; 23,24,25 ; and 24,25,26 will have 24 but next 25,26,27 will not have multiple of 4.. the next three will have 28 and so on
_________________

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

13 Nov 2016, 22:02

fluke wrote:

shashankp27 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

(C-1)C(C+1) should be divisible by 12.

Question is: How many of the integers from 20 to 99 are either ODD or Divisible by 4.

ODD=(99-21)/2+1=40 Divisible by 4= (96-20)/4+1=20 Total=99-20+1=80

P=Favorable/Total=(40+20)/80=60/80=3/4

Ans: "C"

Thanks a lot. This actually taught me how to divide the problem into fragments to understand what exactly is being asked. If I would have sat with all the numbers and trying to make them fit the bill I would have exhausted 10 min on it.

Now when ever I face such critical problem I will apply your strategy.

Status: You have to have the darkness for the dawn to come

Joined: 09 Nov 2012

Posts: 280

Daboo: Sonu

GMAT 1: 590 Q49 V20

GMAT 2: 730 Q50 V38

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

02 Jul 2017, 11:05

Here is the OE

Probability is (favorable outcomes)/(total # of possibilities). There are 99 – 20 + 1 = 80 possible values for c, so the unknown is how many of these c values yield a c3 – c that is divisible by 12. The prime factorization of 12 is 2 × 2 × 3. There are several ways of thinking about this: numbers are divisible by 12 if they are divisible by 3 and by 2 twice, or if they are multiples of both 4 and 3, or if half of the number is an even multiple of 3, etc. The expression involving c can be factored. c3 – c = c(c2 – 1) = c(c – 1)(c + 1) These are consecutive integers. It may help to put them in increasing order: (c – 1)c(c + 1). Thus, this question has a lot to do with Consecutive Integers (Manhattan GMAT Number Properties Strategy Guide), and not only because the integers 20 to 99 themselves are consecutive. In any set of three consecutive integers, a multiple of 3 will be included. Thus, (c – 1)c(c + 1) is always divisible by 3 for any integer c. This takes care of part of the 12. So the question simply becomes “How many of the possible (c – 1)c(c + 1) values are divisible by 4?” Since the prime factors of 4 are 2's, it makes sense to think in terms of Odds and Evens (Manhattan GMAT Number Properties Strategy Guide). (c – 1)c(c + 1) could be (E)(O)(E), which is definitely divisible by 4, because the two evens would each provide at least one separate factor of 2. Thus, c3 – c is divisible by 12 whenever c is odd, which are the cases c = 21, 23, 25, …, 95, 97, 99. That's ((99 – 21)/2) + 1 = (78/2) + 1 = 40 possibilities. Alternatively, (c – 1)c(c + 1) could be (O)(E)(O), which will only be divisible by 4 when the even term itself is a multiple of 4. Thus, c3 – c is also divisible by 12 whenever c is a multiple of 4, which are the cases c = 20, 24, 28, …, 92, 96. That's ((96 – 20)/4) + 1 = (76/4) + 1 = 20 possibilities. The probability is thus (40 + 20)/80 = 60/80 = 3/4.
_________________

You have to have the darkness for the dawn to come.

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

02 Jul 2017, 19:49

hi bunuel, could you please tell me why we are taking total as 80? the question is probability of c^3 - c is divisible by 12, so in the denominator, dont we need the total number of products of consecutive integers in the set 20-99? this number will be 78. could you please tell why are we taking simply the number of elements in the above set?

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

03 Jul 2017, 02:56

kkrrsshh wrote:

hi bunuel, could you please tell me why we are taking total as 80? the question is probability of c^3 - c is divisible by 12, so in the denominator, dont we need the total number of products of consecutive integers in the set 20-99? this number will be 78. could you please tell why are we taking simply the number of elements in the above set?

Not sure I can follow you. What number will be 78? There are total of 80 integers from 20 to 99, inclusive: 20, 21, 22, 23, ..., 98, 99.
_________________

If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

08 Mar 2018, 21:50

Bunuel 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

Two things: 1. There are total of 80 integers from 20 to 99, inclusive: 20, 21, ..., 99. 2. C^3-C=(C-1)*C*(C+1): we have the product of 3 consecutive integers, which is always divisible by 3. So the question basically is whether (C-1)*C*(C+1) is divisible by 4.

Next, the only way the product NOT to be divisible by 4 is C to be even but not a multiple of 4, in this case we would have (C-1)*C*(C+1)=odd*(even not multiple of 4)*odd.

Now, out of first the 4 integers {20, 21, 22, 23} there is only 1 even not multiple of 4: 22. All following groups of 4 will also have only 1 even not multiple of 4 (for example in {24, 25, 26, 27} it's 26, and in {96, 97, 98, 99} it's 98, always 3rd in the group) and as our 80 integers are entirely built with such groups of 4 then the overall probability that C is not divisible by 4 is 1/4. Hence the probability that it is divisible by 4 is 1-1/4=3/4.

Answer: C.

nkhosh wrote:

Would you please explain (1) The formulas you used for finding the # of ODD and "Divisible by 4" outcomes? AND: (2) Why are we looking for the number of ODD integers within 20-99? I'm confused b/c 23 for example is not divisible by 12....

Thank you for the great post

1. There are even # of consecutive integers in our range - 80 (from 20 to 99, inclusive). Out of even number of consecutive integers half is always odd and half is always even, thus numbers of odd integers in the given range is 40.

2. \(# \ of \ multiples \ of \ x \ in \ the \ range =\) \(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

So, for our case as 96 is the last multiple of 4 in the range and 20 is the first multiple of 4 in the range then total # of multiples of 4 in the range is (96-20)/4+1=20.

Or look at it in another way out of 20 consecutive even integers we have half will be even multiples of 4 and half will be even not multiples of 4.

Hope it's clear.

Bunuel kindly explain the grouping method or provide a link to the post where it is explained, My doubt is why you only considered 4 as a group and why not 6 or 12 or 8 ??

Bunuel wrote:

Now, out of first the 4 integers {20, 21, 22, 23} there is only 1 even not multiple of 4: 22. All following groups of 4 will also have only 1 even not multiple of 4 (for example in {24, 25, 26, 27} it's 26, and in {96, 97, 98, 99} it's 98, always 3rd in the group) and as our 80 integers are entirely built with such groups of 4 then the overall probability that C is not divisible by 4 is 1/4. Hence the probability that it is divisible by 4 is 1-1/4=3/4.

_________________

Give Kudos for correct answer and/or if you like the solution.

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

24 Jul 2018, 23:20

I was blind but now I see... thanks for the prompt response!

One follow up question...would you be able to explain the following rule that I've been experimenting with for these types of questions...

Sequence (Denominator): X Numbers (# of products in the numerator): Y 1 - Y/X = X/X - Y/X = (X-Y)/X

For example, in this question, (C-1)(C)(C+1)/12 Sequence: 12 Numbers: 3 1 - 3/12 = 12/12 - 3/12 = 8/12 = 3/4

It appears to work but I don't fully understand why...[/quote]

This is adding/subtracting fractions by finding common denominator:

Please do not get me wrong but you should really brush up the basics before attempting questions.[/quote]

My apologies...I should have worded my question more clearly as that was not what I was intending to ask (I understand how to add/subtract fractions). My intended question relates to the following post from another thread...https://gmatclub.com/forum/if-an-intege ... 22078.html

"3-sec solution:

If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

Sequence of 4 numbers: 4 Numbers we have: 2

Answer: 1-2/4 = 1/2

You can see how this method works also on harder questions letting us solve them in a few seconds. eg. If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

Sequence of 8 numbers: 8 Numbers we have: 3

Answer 1-3/8= 5/8.

Similar questions: if-integer-c-is-randomly-selected-from-20-to-99-inclusive-121561.html if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html"

To be clear, I understand how adding/subtracting fractions work. What I don't understand is the logic/rationale as to why this approach seems to always hold true and yield a correct answer?

If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

Updated on: 29 Jul 2018, 02:19

Bunuel 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

Two things: 1. There are total of 80 integers from 20 to 99, inclusive: 20, 21, ..., 99. 2. C^3-C=(C-1)*C*(C+1): we have the product of 3 consecutive integers, which is always divisible by 3. So the question basically is whether (C-1)*C*(C+1) is divisible by 4.

Next, the only way the product NOT to be divisible by 4 is C to be even but not a multiple of 4, in this case we would have (C-1)*C*(C+1)=odd*(even not multiple of 4)*odd.

Now, out of first the 4 integers {20, 21, 22, 23} there is only 1 even not multiple of 4: 22. All following groups of 4 will also have only 1 even not multiple of 4 (for example in {24, 25, 26, 27} it's 26, and in {96, 97, 98, 99} it's 98, always 3rd in the group) and as our 80 integers are entirely built with such groups of 4 then the overall probability that C is not divisible by 4 is 1/4. Hence the probability that it is divisible by 4 is 1-1/4=3/4.

Answer: C.

nkhosh wrote:

Would you please explain (1) The formulas you used for finding the # of ODD and "Divisible by 4" outcomes? AND: (2) Why are we looking for the number of ODD integers within 20-99? I'm confused b/c 23 for example is not divisible by 12....

Thank you for the great post

1. There are even # of consecutive integers in our range - 80 (from 20 to 99, inclusive). Out of even number of consecutive integers half is always odd and half is always even, thus numbers of odd integers in the given range is 40.

2. \(# \ of \ multiples \ of \ x \ in \ the \ range =\) \(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

So, for our case as 96 is the last multiple of 4 in the range and 20 is the first multiple of 4 in the range then total # of multiples of 4 in the range is (96-20)/4+1=20.

Or look at it in another way out of 20 consecutive even integers we have half will be even multiples of 4 and half will be even not multiples of 4.

Hope it's clear.

Hi Bunuel Thanks for the explanation. I have a doubt in this.

After creating groups, why are we checking for the even integers not divisible by 4. Can't we check directly from the set that how many numbers are divisible by 4?

{20,21,22,23}=> Here only 1 number is divisible by 4 {24,25,26,27}=> Here only 1 number is divisible by 4 {28,29,30,31}=> Here only 1 number is divisible by 4 {32,33,34,35}=> Here only 1 number is divisible by 4

Similarly, {96,97,98,99}=> Here only 1 number is divisible by 4

So p(E) = 1/4

Please tell me where I am wrong. Thanks in advance

Originally posted by suramya26 on 29 Jul 2018, 02:03.
Last edited by suramya26 on 29 Jul 2018, 02:19, edited 1 time in total.

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

29 Jul 2018, 02:15

suramya26 wrote:

Bunuel 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

Two things: 1. There are total of 80 integers from 20 to 99, inclusive: 20, 21, ..., 99. 2. C^3-C=(C-1)*C*(C+1): we have the product of 3 consecutive integers, which is always divisible by 3. So the question basically is whether (C-1)*C*(C+1) is divisible by 4.

Next, the only way the product NOT to be divisible by 4 is C to be even but not a multiple of 4, in this case we would have (C-1)*C*(C+1)=odd*(even not multiple of 4)*odd.

Now, out of first the 4 integers {20, 21, 22, 23} there is only 1 even not multiple of 4: 22. All following groups of 4 will also have only 1 even not multiple of 4 (for example in {24, 25, 26, 27} it's 26, and in {96, 97, 98, 99} it's 98, always 3rd in the group) and as our 80 integers are entirely built with such groups of 4 then the overall probability that C is not divisible by 4 is 1/4. Hence the probability that it is divisible by 4 is 1-1/4=3/4.

Answer: C.

nkhosh wrote:

Would you please explain (1) The formulas you used for finding the # of ODD and "Divisible by 4" outcomes? AND: (2) Why are we looking for the number of ODD integers within 20-99? I'm confused b/c 23 for example is not divisible by 12....

Thank you for the great post

1. There are even # of consecutive integers in our range - 80 (from 20 to 99, inclusive). Out of even number of consecutive integers half is always odd and half is always even, thus numbers of odd integers in the given range is 40.

2. \(# \ of \ multiples \ of \ x \ in \ the \ range =\) \(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

So, for our case as 96 is the last multiple of 4 in the range and 20 is the first multiple of 4 in the range then total # of multiples of 4 in the range is (96-20)/4+1=20.

Or look at it in another way out of 20 consecutive even integers we have half will be even multiples of 4 and half will be even not multiples of 4.

Hope it's clear.

Hi Bunuel Thanks for the explanation. I have a doubt in this.

After creating groups, why are we checking for the even integers not divisible by 4. Can't we check directly from the set that how many numbers are divisible by 4?

{20,21,22,24}=> Here only 1 number is divisible by 4 {25,26,27,28}=> Here only 1 number is divisible by 4 {29,30,31,32}=> Here only 1 number is divisible by 4 {33,34,35,36}=> Here only 1 number is divisible by 4

Similarly, {96,97,98,99}=> Here only 1 number is divisible by 4

So p(E) = 1/4

Please tell me where I am wrong. Thanks in advance

Yes, in {20,21,22,23} only one number is divisible by 4 but (C-1)*C*(C+1) is divisible by 12 also when C = 21 or 23. So, for three numbers out of {20,21,22,23} (C-1)*C*(C+1) IS divisible by 12, for 20, 21, and 23. Similarly for all other sets of four numbers there will be 3 out of 4 numbers for which (C-1)*C*(C+1) will be divisible by 12. P = 3/4.

Re: If integer C is randomly selected from 20 to 99, inclusive.
[#permalink]

Show Tags

29 Jul 2018, 02:26

Bunuel 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

Two things: 1. There are total of 80 integers from 20 to 99, inclusive: 20, 21, ..., 99. 2. C^3-C=(C-1)*C*(C+1): we have the product of 3 consecutive integers, which is always divisible by 3. So the question basically is whether (C-1)*C*(C+1) is divisible by 4.

Next, the only way the product NOT to be divisible by 4 is C to be even but not a multiple of 4, in this case we would have (C-1)*C*(C+1)=odd*(even not multiple of 4)*odd.

Now, out of first the 4 integers {20, 21, 22, 23} there is only 1 even not multiple of 4: 22. All following groups of 4 will also have only 1 even not multiple of 4 (for example in {24, 25, 26, 27} it's 26, and in {96, 97, 98, 99} it's 98, always 3rd in the group) and as our 80 integers are entirely built with such groups of 4 then the overall probability that C is not divisible by 4 is 1/4. Hence the probability that it is divisible by 4 is 1-1/4=3/4.

Answer: C.

nkhosh wrote:

Would you please explain (1) The formulas you used for finding the # of ODD and "Divisible by 4" outcomes? AND: (2) Why are we looking for the number of ODD integers within 20-99? I'm confused b/c 23 for example is not divisible by 12....

Thank you for the great post

1. There are even # of consecutive integers in our range - 80 (from 20 to 99, inclusive). Out of even number of consecutive integers half is always odd and half is always even, thus numbers of odd integers in the given range is 40.

2. \(# \ of \ multiples \ of \ x \ in \ the \ range =\) \(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

So, for our case as 96 is the last multiple of 4 in the range and 20 is the first multiple of 4 in the range then total # of multiples of 4 in the range is (96-20)/4+1=20.

Or look at it in another way out of 20 consecutive even integers we have half will be even multiples of 4 and half will be even not multiples of 4.

Hope it's clear.

Hi Bunuel Thanks for the explanation. I have a doubt in this.

After creating groups, why are we checking for the even integers not divisible by 4. Can't we check directly from the set that how many numbers are divisible by 4?

{20,21,22,24}=> Here only 1 number is divisible by 4 {25,26,27,28}=> Here only 1 number is divisible by 4 {29,30,31,32}=> Here only 1 number is divisible by 4 {33,34,35,36}=> Here only 1 number is divisible by 4

Similarly, {96,97,98,99}=> Here only 1 number is divisible by 4

So p(E) = 1/4

Please tell me where I am wrong. Thanks in advance[/quote]

Yes, in {20,21,22,23} only one number is divisible by 4 but (C-1)*C*(C+1) is divisible by 12 also when C = 21 or 23. So, for three numbers out of {20,21,22,23} (C-1)*C*(C+1) IS divisible by 12, for 20, 21, and 23. Similarly for all other sets of four numbers there will be 3 out of 4 numbers for which (C-1)*C*(C+1) will be divisible by 12. P = 3/4.