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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. 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
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Originally posted by shashankp27 on 05 Oct 2011, 09:44.
Last edited by Bunuel on 07 Jul 2013, 06:27, edited 2 times in total.
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Re: divisible by 12 Probability
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15 Jan 2012, 17:50
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.
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Re: divisible by 12 Probability
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05 Oct 2011, 10:15
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"
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Re: divisible by 12 Probability
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08 Oct 2011, 01:55
How many of the integers from 20 to 99 are either ODD or Divisible by 4. could you explain why have u take "either ODD" Thank you
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Re: divisible by 12 Probability
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08 Oct 2011, 02:06
gaurav2k101 wrote: How many of the integers from 20 to 99 are either ODD or Divisible by 4. could you explain why have u take "either ODD"
Thank you We know C^3C=(C1)C(C+1), product of 3 consecutive integers. Now, 19,20,21 20,21,22 21,22,23 22,23,24 23,24,25 ... 96,97,98 97,98,99 98,99,100 These are the entire set. Note; the middle value represents "C" and we know 20<=C<=99; (C1) AND (C+1) are left and right values based on C. Total count=9920+1=80 We need to see how many of these sets will be divisible by 4; If C=odd; left value=C1=even; C+1=even; Multiplication of 2 Even numbers will always be divisible by 4 because it will contain at least two 2's in its factors. e.g. 48,49,50 49=odd 48=even 50=even; So, 48*49*50 must be a multiple of 4 as it has 2 even numbers. Thus, we count all odds C's, for it will make the left and right values even. Or, 23, 24, 25 Since, 24 is a multiple of 4, it will also be divisible by 4. Thus, we count that too. But, sets such as 21,22,23: will not be divisible by 4 as 21 AND 23 are odds and don't contain any 2 in their factors. and 22 is not divisible by 4. forgot to mention: the product of three consecutive integers will always be divisible by 3, so we don't need to mind that. We just need to make sure that there are at least 2 2's.
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Re: divisible by 12 Probability
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15 Jan 2012, 17:00
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" 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



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Re: divisible by 12 Probability
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15 Jan 2012, 18:26
Thank you very much for the explanation! It definitely clarified the concept
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Re: If integer C is randomly selected from 20 to 99, inclusive
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27 May 2013, 08:26
can v try and solve the problem in the foll. way: C^3C = C(C^21)/12 so then v could check the probability for C/12. Now between 20 and 99 there are 7 integers i.e 24, 36, 48, 60, 72, 84, 96. I don't know how to proceed further.



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Re: divisible by 12 Probability
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07 Jul 2013, 08:54
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. Hi Bunnel, Is the below method correct: As per the 30 sec approach: (c1)c(c+1) The initial 5 entries (before the multiple of 12) have 4 entries which have a multiple of 12 19.20.21 20.21.22 21.22.23 22.23.24 23.24.25
Then their are 8 groups of 12, and each having a probability of being divisible by 12 = 9/12=3/4 24.25.26 36.37.38 25 37 26 . . 35.36.37 47.48.49..............8 such groups.. In the last 3 entries 2 are divisible by 12 96.97.98 97.98.99 98.99.100 So among the first 5 and last 3  (4+2)/(5+3)=6/6=3/4 Thus the overall probability is 3/4.



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Re: divisible by 12 Probability
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13 Sep 2013, 03:28
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 (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.
Answer: C. Hi Bunuel, Product of 3 consecutive integers will be always NOT ONLY divisible by 3 BUT ALSO divisible by 2...Right? Then,can we rephrase this question as whether (C1)*C*(C+1) is divisible by 2? Please clarify.
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Re: If integer C is randomly selected from 20 to 99, inclusive.
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13 Sep 2013, 04:38
My approach was just two steps: 1) As we have (C1)*C*(C+1) we can forget about devisibility by 3 and think only of div. by 4; 2) Then I didn't care about the total number of sets but considered that moving through number line and taking triples of sequent numbers we have only one option of four possible not to cover multiple of 4. So, the answer must be 11/4 = 3/4.
I realize that it could be less straightforward if number of possible options was not 80, but in that case denominator of the answer would be smth less simple than 2,3,4,5 as we see in answers.



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Re: divisible by 12 Probability
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13 Sep 2013, 04:43
bagdbmba wrote: 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 (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.
Answer: C. Hi Bunuel, Product of 3 consecutive integers will be always NOT ONLY divisible by 3 BUT ALSO divisible by 2...Right? Then,can we rephrase this question as whether (C1)*C*(C+1) is divisible by 2? Please clarify. I'm quite sure this works only with prime numbers. Otherwise you can prove divisibility of 2 by 2^N =)



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Re: If integer C is randomly selected from 20 to 99, inclusive.
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08 Mar 2014, 11:04
C^3C = C(C1)(C+1)
Divisibility by 12 requires divisibility by both 4 and 3.
given C1, C and C+1 are 3 consec. integers, divisibility by 3 is guaranteed. The question remains whether it is divisible by 4.
2 cases: a) C is odd > C1 and C+1 are even which guarantees at least 2 factors of 2 and hence divisible by 4 b) C is even > only C is even > divisibility by 4 requires C to be divisible by 4
2099 inclusive is 80 integers, 40 odd and 40 even. Of those that are even every other one is divisible by 4, that is half of the 40. Therefore there are 20 divisible by 4
Thus the probability is (40+20)/80 = 3/4



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



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Re: If integer C is randomly selected from 20 to 99, inclusive.
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Re: If integer C is randomly selected from 20 to 99, inclusive.
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22 Nov 2014, 03:06
Ah! I had to take some time to figure out this 1. This working might be helpful to you! C^3  C= (c1)(c)(c+1) > this is some rule Any integer (c1)(c)(c+1) will be divisible by 12, if any of these numbers are factors of 4 and 3, either alone or together. Now, in a set of any 3 consecutive integers, one will automatically be divisible by 3. So the Q is simplied as whether the integer is a multiple of 4! Now, there are 2 possibilities  C is even or odd. a) C is even. In this case 2 numbers (c1 and c+1) will be odd. This leads to 2 sub possibilities! (i) C is a multiple of 4  This means the integer will be divisible by 4. (We will have: odd X multiple of 4 X odd) (ii) C is not a multiple of 4 This means the integer will NOT be divisible by 4. (We will have: odd X non multiple of 4 X odd) b) C is odd. two numbers (c1 and c+1) are divisible by 2 and therefore the set is divisble by 4. Now the total numbers in the set are 80 (not 79, since 20 and 99 are included). The number of possibilities for C are (a) (i) > 99/4  19/4 = 24 4 = 20 (a) (ii) > 99/2  19/2 = 49 9 = 40; reduce 20 from (a)(i) ==> 20 (b) > 100/2  20/2 = 50 10 = 40 Sum of (a) (i) + (a) (ii) + (b) > 20+20+40 = 80 (80 is the total number of numbers between 20 and 99, this means we have considered all possibilities ) Favourable outcomes above (a) (i) + (b) = 60Therefore probability = 60/80 or 3/4 Answer = C
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Re: If integer C is randomly selected from 20 to 99, inclusive.
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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 (C1)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 (C1)C(C+1) is divisible by 4. We've 80 numbers b/w 2099, 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.



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Re: If integer C is randomly selected from 20 to 99, inclusive.
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28 Jun 2015, 02:25
After watching the answer posted for this question, i would like to point out a method which requires least calculation and time. Given expression > (C1)*C*(C+1) is always divisible by 3 (Product of three consecutive numbers) Now the question here is to whether this will be divisible by 4 ? This can be solved by using probability of selecting 3 numbers out of 4 numbers whose product is divisible by 4 where numbers are consecutive is 3/4



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Re: If integer C is randomly selected from 20 to 99, inclusive.
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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 (c1) 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....



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If integer C is randomly selected from 20 to 99, inclusive.
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28 Jun 2015, 05:10
matvan 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 The answer is C My solution is longer, time consuming. The given expression is (c1) 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.... Where I say that induction method is the soul of Math's we should not forget that Induction methods with a little bit of Observation is the finest way to solve the question. The key is "To question how and when will (c1) c (c+1) be divisible by 12"12 = 4*3 i.e. (c1) c (c+1) will be divisible by 12 if it's divisible by 4 as well by 3 in order to make sure that this is divisible by 4 1) either (c1) should be divisible by 2 so that (c+1) is also even and their product is divisible by 4 i.e. (c1) can b any even number from 20 through 99 = (98/2)(18/2) = 499 = 40 cases2) or if (c1) is odd then (c+1)will also be odd i.e. c must be the multiple of 4 itself i.e. c can be any multiple of 4 from 20 though 99 = (99/4)(19/4) = 244 = 20 casesTotal favorable cases = 40 + 20 = 60Hence, Probability = 60/80 = 3/4I hope this helps!
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