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If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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29 Jul 2015, 17:04
There are 80 numbers from 2099 inclusive.
c^3  c = 12k c(c^21) = c(c+1)(c1) = 12k = 2*2*3*k = 4*3*k Make c = 4, then c1 = 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.



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If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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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 :\(last = first + (n1)*1\) => we get: 99 = 20 + n1 => n = 9920 +1 = 80 2. \(C^3  C = (C1)*C*(C+1)\) => 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 \((C1)*C*(C+1) = (2q)(2q+1)(2q+2) = q(2q+1)(q+1)*4\) 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 : \(Last = First + (n1)*2\), we'll find that there are 40 odd. 5. The C is even 5.1. When C is even we've \((C1)*C*(C+1) = (2q1)(2q)(2q+1)\)=> 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
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Re: If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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13 Apr 2016, 00:00
Bunuel wrote: If integer C is randomly selected from 20 to 99, inclusive. What is the probability that c^3c 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^3C=(C1)*C*(C+1): we have the product of 3 consecutive integers, which is always divisible by 3. So the question basically is whether (C1)*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 (C1)*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 (f or 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 11/4=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 2099? 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 (9620)/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 (c1)*c*(c+1) will be divisible by 12. For ex: 48 has 47 and 49 as consecutive integers when 47 is substituted> (471)*47*(47+1) is divisible by twelve. same as 49> (491)*49*(49+1) is divisible by 12. Total their are 21 integers as above so probability = 21/80 ~20/80 =1/4.



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Re: If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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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 (c1)*c*(c+1) will be divisible by 12. For ex: 48 has 47 and 49 as consecutive integers when 47 is substituted> (471)*47*(47+1) is divisible by twelve. same as 49> (491)*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
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If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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17 Jul 2016, 04:29
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 Question: is What is the probability that\(\frac{C^3  C}{12}\) >\(\frac{C(C^21)}{4*3}\) >\(\frac{(C1)C(C+1)}{4*3}\)? (20,21,22,23),(24,25,26,27),(28,29,30,31),......,(96,97,98,99) In above groups group of three numbers divisible by 12 and the group always excluded one number that is not divisible by 4.( any three consecutive numbers are always divisible by 3) So the probability is \(\frac{3}{4}\) Correct Answer C
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Re: If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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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 (C1)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=(9921)/2+1=40 Divisible by 4= (9620)/4+1=20 Total=9920+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.



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Re: If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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01 Mar 2017, 17:59
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Re: If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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07 Mar 2017, 17:44
Quote: 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 We are given that an integer C is to be chosen at random from the integers 20 to 99 inclusive, and we need to determine the probability that C3 – C will be divisible by 12. We should recall that when a number is divisible by 12, it is divisible by 4 and 3. We should also recognize that C3 – C = C(C2 – 1) = C(C – 1)(C + 1) = (C – 1)(C)(C + 1) is a product of three consecutive integers. Furthermore, we should recognize that any product of three consecutive integers is divisible by 3! = 6; thus, it’s divisible by 3. We have to make sure it’s also divisible by 4. Case 1: C is odd If C is odd, then both C – 1 and C + 1 will be even. Moreover, either C – 1 or C + 1 will be divisible by 4. Since C(C – 1)(C + 1) is already divisible by 3 and now we know it’s also divisible by 4, C(C – 1)(C + 1) will be divisible by 12. Since there are 80 integers between 20 and 99 inclusive, and half of those integers are odd, there are 40 odd integers (i.e., 21, 23, 25, …, 99) from 20 to 99 inclusive. Thus, when C is odd, there are 40 instances in which C(C – 1)(C + 1) will be divisible by 12. Case 2: C is even If C is even, the both C – 1 and C + 1 will be odd. If C is a multiple of 4 but not a multiple of 3, then either C – 1 or C + 1 will be divisible by 3 (for example, if C = 28, then C – 1 = 27 is divisible by 3, and if C = 44, then C + 1 = 45 is divisible by 3). In this case, C(C – 1)(C + 1) will be divisible by 12. If C is a multiple 4 and also a multiple of 3, then C is a multiple of 12 and of course C(C – 1)(C + 1) will be divisible by 12. Therefore, if C is even and a multiple of 4, then C(C – 1)(C + 1) will be divisible by 12. So, let’s determine the number of multiples of 4 between 20 and 99 inclusive. Number of multiples of 4 = (96 – 20)/4 + 1 = 76/4 + 1 = 20. Thus, when C is even, there are 20 instances in which C(C – 1)(C + 1) will be divisible by 12. In total, there are 40 + 20 = 60 outcomes in which C(C – 1)(C + 1) will be divisible by 12. Thus, the probability that C(C – 1)(C + 1) will be divisible by 12 is: 60/80 = 6/8 = 3/4. Answer: C
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Re: If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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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.
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Re: If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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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 2099? this number will be 78. could you please tell why are we taking simply the number of elements in the above set?



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Re: If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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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 2099? 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.
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If integer C is randomly selected from 20 to 99, inclusive. [#permalink]
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08 Mar 2018, 21:50
Bunuel wrote: If integer C is randomly selected from 20 to 99, inclusive. What is the probability that c^3c 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^3C=(C1)*C*(C+1): we have the product of 3 consecutive integers, which is always divisible by 3. So the question basically is whether (C1)*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 (C1)*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 (f or 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 11/4=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 2099? 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 (9620)/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 11/4=3/4.
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