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If 1=<n<=99, what is the probability that n(n + 1) is perfec

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If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink] New post 11 Dec 2010, 22:52
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If 1<n<99, what is the probability that n(n + 1) is perfectly divisible by 3 ?

[Reveal] Spoiler:
For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.

Considering n is a multiple of 3:
Between 1 to 99, the number of multiples of 3 = (99 - 3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]

Considering (n + 1) is a multiple of 3:
Between 1 to 99, the number of multiples of 3 would be from 2-98 and would also be 33 multiples.

Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd)
= 33/99 + 33/99 = 66/99 = 22/33 = 2/3


I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble
(nL(last)-nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?
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Re: Beginner's Forum Question [#permalink] New post 11 Dec 2010, 23:08
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this formula is in the MGMAT math book..

really useful to know... you add 1 as you generally underestimate by 1 at the extreme -- this evenly spaced sets formula is the basis for the simpler formula: (last-First) + 1 when you have to count out the number of elements in a consecutive sequence 1, 2, 3....
You are basically doing (last-first)/1 + 1 ---> you are basically dividing by 1 there as that is the common space b/w the elements...

Knowing that you can apply this formula for evenly spaced sets should be enough...
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Re: Beginner's Forum Question [#permalink] New post 11 Dec 2010, 23:22
thanks for the quick reply! kudos for the help.
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Re: Beginner's Forum Question [#permalink] New post 12 Dec 2010, 00:17
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mmcooley33 wrote:
If 1=<n<=99, what is the probability that n(n + 1) is perfectly divisible by 3 ?

For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.

Considering n is a multiple of 3:
Between 1 to 99, the number of multiples of 3 = (99 - 3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]

Considering (n + 1) is a multiple of 3:
Between 1 to 99, the number of multiples of 3 would be from 2-98 and would also be 33 multiples.

Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd)
= 33/99 + 33/99 = 66/99 = 22/33 = 2/3


I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble
(nL(last)-nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?


First about the multiple of x in the given range:

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\), (check this: totally-basic-94862.html?hilit=multiple%20range).

So, there will be \(\frac{99-3}{3}+1=33\) multiples of 3 in the range from 1 to 99, inclusive: 3, 6, 9, ..., 99;

n+1 to be a multiple of 3 in the range from 1 to 99, inclusive n must be 2, 5, 8, ..., 98, one less than above values, so also 33 options for n;

Hence, n(n+1) to be a multiple of 3 n can take 33+33=66 values: P=(favorable outcomes)/(total # of outcomes)=66/99=2/3.

30 second approach:

Take first 3 numbers from the given range: 1, 2, and 3 in order n(n+1) to be a multiple of 3 n can be either 2 or 3 P=2/3. For next 3 numbers the probability will be the same and as 99 equals to 3*33 (similar 33 blocks of 3) then for complete range the probability will still be 2/3.

Answer: 2/3.

Hope it helps.
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Re: Beginner's Forum Question [#permalink] New post 13 Dec 2010, 07:41
Ah, that 30 sec approach looks good.
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Re: Beginner's Forum Question [#permalink] New post 07 Jul 2013, 03:12
Bunuel wrote:
mmcooley33 wrote:
If 1=<n<=99, what is the probability that n(n + 1) is perfectly divisible by 3 ?

For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.

Considering n is a multiple of 3:
Between 1 to 99, the number of multiples of 3 = (99 - 3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]

Considering (n + 1) is a multiple of 3:
Between 1 to 99, the number of multiples of 3 would be from 2-98 and would also be 33 multiples.

Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd)
= 33/99 + 33/99 = 66/99 = 22/33 = 2/3


I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble
(nL(last)-nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?


First about the multiple of x in the given range:

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\), (check this: totally-basic-94862.html?hilit=multiple%20range).

So, there will be \(\frac{99-3}{3}+1=33\) multiples of 3 in the range from 1 to 99, inclusive: 3, 6, 9, ..., 99;

n+1 to be a multiple of 3 in the range from 1 to 99, inclusive n must be 2, 5, 8, ..., 98, one less than above values, so also 33 options for n;

Hence, n(n+1) to be a multiple of 3 n can take 33+33=66 values: P=(favorable outcomes)/(total # of outcomes)=66/99=2/3.

30 second approach:

Take first 3 numbers from the given range: 1, 2, and 3 in order n(n+1) to be a multiple of 3 n can be either 2 or 3 P=2/3. For next 3 numbers the probability will be the same and as 99 equals to 3*33 (similar 33 blocks of 3) then for complete range the probability will still be 2/3.

Answer: 2/3.

Hope it helps.



Bump for the 30 sec approach.
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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink] New post 07 Jul 2013, 05:36
Expert's post
mmcooley33 wrote:
If 1<n<99, what is the probability that n(n + 1) is perfectly divisible by 3 ?

[Reveal] Spoiler:
For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.

Considering n is a multiple of 3:
Between 1 to 99, the number of multiples of 3 = (99 - 3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]

Considering (n + 1) is a multiple of 3:
Between 1 to 99, the number of multiples of 3 would be from 2-98 and would also be 33 multiples.

Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd)
= 33/99 + 33/99 = 66/99 = 22/33 = 2/3


I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble
(nL(last)-nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?


Similar questions to practice:
if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html
if-integer-c-is-randomly-selected-from-20-to-99-inclusive-121561.html
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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink] New post 09 Jul 2014, 06:29
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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink] New post 22 Nov 2014, 02:19
Wow! I did this one very fast and correctly! :D (After getting a series of other questions in this type wrong)

Here is a very simple solution. For n(n+1) to be divisible by 3, either
a) n must be a multiple of 3 --> Number of possibilities = 99/3 = 33 OR
b) (n -1) must be a multiple of 3 --> Number of possibilities = 99/3 = 33

Now, favorable possibilities = 66 and total possibilities = 100 (100 because 99 and 1 are both included).

Therefore, probability = 66/100 or 2/3 :D

There's a discussion going on here for this question: an-integer-n-between-1-and-99-inclusive-is-to-be-chosen-at-160998.html#p1445924
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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec   [#permalink] 22 Nov 2014, 02:19
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