If 1=

Question

If  1 \leq n \leq 99 , what is the probability that n(n + 1) is divisible by 3 ?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

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

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?

Bunuel · Accepted Answer

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.

gmat1011 · Answer

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...

mmcooley33 · Answer

thanks for the quick reply! kudos for the help.

tfincham86 · Answer

Ah, that 30 sec approach looks good.

cumulonimbus · Answer

Bump for the 30 sec approach.

Bunuel · Answer

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

AdmitJA · Answer

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

ScottTargetTestPrep · Answer

We are given that an integer n is to be selected at random from 1 to 99 inclusive, and we must determine the probability that n(n+1) will be divisible by 3.
 
Since probability = favorable outcomes/total outcomes and we know that the total number of outcomes is 99, because there are 99 integers from 1 to 99 inclusive, we need to determine the number of values of n such that n(n+1) is divisible by 3.
 
First, we can determine the number of values of n that are divisible by 3, that is, the number of multiples of 3 that are between 3 and 99 inclusive. To calculate this, we can use the formula:
 
(Largest multiple of 3 – smallest multiple of 3)/3 + 1
 
(99 - 3)/3 + 1
 
96/3 + 1
 
32 + 1 = 33
 
Thus, there are 33 multiples of 3 between 1 and 99 inclusive. That is, the value of n can be any one of these 33 multiples of 3 so that n(n + 1) will be divisible by 3.

Similarly, if n + 1 is a multiple of 3, n(n + 1) will be also be divisible by 3. Since we know that there are 33 values of n that are multiples of 3, there must be another 33 values of n such that n + 1 is a multiple of 3. Let’s expand on this idea:
 
When n = 2, n + 1 = 3, and thus n(n+1) is a multiple of 3.

When n = 23, n + 1 = 24, and thus n(n+1) is a multiple of 3.
 
When n = 98, n + 1 = 99, and thus n(n+1) is a multiple of 3.
 
We can see that there are 33 values of n that are multiples of 3, and 33 more values of n for n + 1 to be multiples of 3. Thus, the probability of selecting a value of n so that n(n+1) is a multiple of 3 is:
 
66/99 = 2/3

cavana · Answer

1) n(n+1) is divisible by 3 when either n is divisible by 3 or (n+1) is divisible by 3.
2) n is divisible by 3 in 99/3=33 cases
3) n+1 is divisible by 3 in 99/3=33 cases
4) 33+33=66
5) 66/99=2/3

Kushchokhani · Answer

Bunuel

The question does not mention that n is an integer. Then can we still solve the problem?

ArnauG · Answer

To determine the probability that n(n + 1) is divisible by 3, we need to consider the different values of n within the given range.

There are three cases to consider:

Case 1: n is divisible by 3.
In this case, n(n + 1) is definitely divisible by 3.

Case 2: n + 1 is divisible by 3.
In this case, n is two less than a multiple of 3, i.e., n = 2, 5, 8, ..., 98. For each of these values, n(n + 1) is divisible by 3.

Case 3: Neither n nor n + 1 is divisible by 3.
In this case, n is one less than a multiple of 3, i.e., n = 1, 4, 7, ..., 97. For each of these values, n(n + 1) is not divisible by 3.

Now, let's count the number of favorable outcomes:

Case 1: There are 33 multiples of 3 between 1 and 99, inclusive.

Case 2: There are 33 values of n in the range 2 to 98, inclusive, where n + 1 is divisible by 3.

Therefore, the total number of favorable outcomes is 33 + 33 = 66.

The total number of possible outcomes is 99 (since 1 ≤ n ≤ 99).

Therefore, the probability that n(n + 1) is divisible by 3 is 66/99, which simplifies to (D) 2/3.

bumpbot · Answer

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

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