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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?

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...

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.

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.

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?

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

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09 Jul 2014, 07:29

Hello from the GMAT Club BumpBot!

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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

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22 Nov 2014, 03:19

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).

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec [#permalink]

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13 Dec 2015, 15:05

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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