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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 [#permalink]  05 Oct 2011, 09:44
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33% (01:54) correct 66% (01:37) wrong based on 6 sessions
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
[Reveal] Spoiler: OA

Last edited by shashankp27 on 06 Oct 2011, 12:21, edited 1 time in total.
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Re: divisible by 12 Probability [#permalink]  05 Oct 2011, 10:15
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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"
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Re: divisible by 12 Probability [#permalink]  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 [#permalink]  08 Oct 2011, 02:06
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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^3-C=(C-1)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; (C-1) AND (C+1) are left and right values based on C.

Total count=99-20+1=80

We need to see how many of these sets will be divisible by 4;

If C=odd; left value=C-1=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 [#permalink]  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

(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"

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
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Re: divisible by 12 Probability [#permalink]  15 Jan 2012, 17:50
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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

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.

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.
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Re: divisible by 12 Probability [#permalink]  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 [#permalink]  22 May 2013, 04:00
Bumping for review and further discussion.
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Re: If integer C is randomly selected from 20 to 99, inclusive   [#permalink] 22 May 2013, 04:00
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