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Intern  Joined: 31 Oct 2010
Posts: 28
If 1=<n<=99, what is the probability that n(n + 1) is perfec  [#permalink]

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2
16 00:00

Difficulty:   45% (medium)

Question Stats: 38% (01:38) correct 63% (01:38) wrong based on 81 sessions

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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 [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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58006
Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec  [#permalink]

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12
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.

Hope it helps.
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##### General Discussion
Manager  Joined: 11 Jul 2010
Posts: 171
Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec  [#permalink]

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1
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...
Intern  Joined: 31 Oct 2010
Posts: 28
Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec  [#permalink]

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thanks for the quick reply! kudos for the help.
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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec  [#permalink]

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Ah, that 30 sec approach looks good.
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Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec  [#permalink]

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1
1
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.

Hope it helps.

Bump for the 30 sec approach.
Math Expert V
Joined: 02 Sep 2009
Posts: 58006
Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec  [#permalink]

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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?

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]

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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]

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mmcooley33 wrote:
If 1<n<99, what is the probability that n(n + 1) is perfectly divisible by 3 ?

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

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

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_________________ Re: If 1=<n<=99, what is the probability that n(n + 1) is   [#permalink] 31 Jul 2019, 01:45
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