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by karthik1239 » Mon Jul 23, 2012 1:21 pm
4 ) 2/3
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by tisrar02 » Mon Jul 23, 2012 5:55 pm
2/3 would be my answer
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by GMATGuruNY » Tue Jul 24, 2012 10:27 am
grandh01 wrote:If n is a positive integer between 1 and 99, inclusive, what is the probability that n(n+1) is a multiple of 3?

1) 1/4
2) 1/3
3) 1/2
4) 2/3
5) 5/6
For n(n+1) to be a multiple of 3, either n or n+1 must be a multiple of 3.

n, n+1, and n+2 are 3 consecutive integers.
Of every 3 consecutive integers, exactly ONE is a multiple of 3.
Thus, P(n+2 is a multiple of 3) = 1/3.
Thus, P(either n or n+1 is a multiple of 3) = 2/3.

The correct answer is D.

Another approach is to WRITE IT OUT and LOOK FOR A PATTERN:
1*2
2*3
3*4
4*5
5*6
6*7
7*8
8*9
9*10
10*11
11*12
12*13
And so on.

The products in red show that, in 2 of every 3 cases, n(n+1) is a multiple of 3.
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by kalpita123 » Tue Jul 24, 2012 1:24 pm
GMATGuruNY wrote:
grandh01 wrote:If n is a positive integer between 1 and 99, inclusive, what is the probability that n(n+1) is a multiple of 3?

1) 1/4
2) 1/3
3) 1/2
4) 2/3
5) 5/6
For n(n+1) to be a multiple of 3, either n or n+1 must be a multiple of 3.

n, n+1, and n+2 are 3 consecutive integers.
Of every 3 consecutive integers, exactly ONE is a multiple of 3.
Thus, P(n+2 is a multiple of 3) = 1/3.
Thus, P(either n or n+1 is a multiple of 3) = 2/3.

The correct answer is D.

Another approach is to WRITE IT OUT and LOOK FOR A PATTERN:
1*2
2*3
3*4
4*5
5*6
6*7
7*8
8*9
9*10
10*11
11*12
12*13
And so on.

The products in red show that, in 2 of every 3 cases, n(n+1) is a multiple of 3.
Hi Mitch,

Could you please solve the below problem with the former method (your first approach to the previous problem) ? I am able to do this with the second approach you have mentioned.

If n is an integer from 1 to 96 (inclusive), what is the probability that n*(n+1)*(n+2) is divisible by 8?

A.1/4
B.1/2
C.5/8
D.7/8
E.3/4

ans- C
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by GMATGuruNY » Tue Jul 24, 2012 1:37 pm
kalpita123 wrote:
Hi Mitch,

Could you please solve the below problem with the former method (your first approach to the previous problem) ? I am able to do this with the second approach you have mentioned.

If n is an integer from 1 to 96 (inclusive), what is the probability that n*(n+1)*(n+2) is divisible by 8?

A.1/4
B.1/2
C.5/8
D.7/8
E.3/4
Case 1: n(n+1)(n+2) = even*odd*even = multiple of 8:
Since every other even integer is a multiple of 4, the product here will always include an even integer and a multiple of 4, resulting in a multiple of 8.
Thus, n can be any even integer between 1 and 96.
96/2 = 48 favorable choices for n.

Case 2: n+1 is a multiple of 8:
The product will be a multiple of 8 if n+1 is a multiple of 8.
Number of multiples of 8 between 1 and 96 = 96/8 = 12.
Thus, there are 12 favorable choices for n+1, implying that there are 12 more favorable choices for n.

Total favorable choices for n = 48+12 = 60.
Favorable choices/Total choices = 60/96 = 5/8.

The correct answer is C.
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by eagleeye » Tue Jul 24, 2012 5:38 pm
grandh01 wrote:If n is a positive integer between 1 and 99, inclusive, what is the probability that n(n+1) is a multiple of 3?
This is how I typically do them.
Step 1: Check that total number is a multiple of 3. (99-1+1 = 99) is. We are good.
Step 2: Since we need to check 3, write down the first 3 numbers.

1.2.3.
Step 3: Just count the pairs ( n*(n+1) )starting with each number and see how many are divisible by 3.
Here the pairs are:
1*2 - no.
2*3 - yes.
3*next number - yes.

So we get 2 out of 3 pairs working. Probability required =2/3.
If n is an integer from 1 to 96 (inclusive), what is the probability that n*(n+1)*(n+2) is divisible by 8
Here again, repeat the same steps.
Step 1: Check the total number is a multiple of 8. (96-1+1 = 96 = 12*8). It is. We are good.
Step 2: Write down the first 8 numbers starting from the first number (In this case it is 1, since 1 to 96 inclusive is our working part).

1.2.3.4.5.6.7.8.

Step 3: Count the triplets ( n*(n+1)*(n+2) ) starting with the first number.

1*2*3 - no.
2*3*4 - yes.
3*4*5 - no.
4*5*6 - yes.
5*6*7 - no.
6*7*8 - yes.
7*8*9 - yes.
8*9*10 - yes.

So we get 5 out of 8 triplets working. Required probability = 5/8.

Cheers!
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by armand_h » Thu Jul 26, 2012 1:44 am
* Probability of n(n+1) being multipole of 3
It's the probability of n being multiple of 3 or n+1 being a multiple of 3
=P(n)+P(n+1)=2*P(n)

P(n)=1/3 (out of 3 consequitive integers one is multiple of 3)

The answer is 2/3

* Probability of n(n+1)(n+2) being multiple of 8
It's the probability that n is multiple of 2 P2(n)
or
n+1 is multiple of 8 P8(n+1)

P2(n)=1/2
P8(n+1)=1/8

1/2+1/8=5/8

Note that if n is an even number, that n(n+2) will always be a multiple of 8
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by info2 » Tue Jun 07, 2016 10:16 am
Hi

In the question n(n+1)(n+2) why don't we consider the overlapping probability. when both events can happen together we subtract the overlapping probability. In this question a number can both be an even number and a multiple of 8. I am a bit confused why dont we solve it that way.?
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by DavidG@VeritasPrep » Tue Jun 07, 2016 10:57 am
info2 wrote:Hi

In the question n(n+1)(n+2) why don't we consider the overlapping probability. when both events can happen together we subtract the overlapping probability. In this question a number can both be an even number and a multiple of 8. I am a bit confused why dont we solve it that way.?
If you're referring to Mitch's approach, there's no overlap. In the first case, n is EVEN. In the second case, n + 1 is a multiple of 8. Anytime n + 1 is a multiple of 8: 8, 16, 24..., n will have to be ODD (7, 15, 23,...)
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by Matt@VeritasPrep » Tue Jun 07, 2016 11:09 pm
info2 wrote:Hi

In the question n(n+1)(n+2) why don't we consider the overlapping probability. when both events can happen together we subtract the overlapping probability. In this question a number can both be an even number and a multiple of 8. I am a bit confused why dont we solve it that way.?
It can, but we're only considering

n even

OR

(n + 1) a multiple of 8

and those CAN'T overlap, since if n is even, (n + 1) must be odd and hence not a multiple of 8.

You're thinking the right way, though: overlapping sets are critical (and common) in probability in general.
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by [email protected] » Wed Jun 08, 2016 2:51 pm
Hi All,

Sequence prompts on the GMAT will follow type of pattern; your ability to figure out that pattern might be essential to the question (such as this question) or helpful in finding an easy approach to answering whatever specific question that the prompt asks for.

You can figure out the first few numbers in the sequence pretty easily...

N=1; 1(1+1) is NOT a multiple of 3
N=2; 2(2+1) IS a multiple of 3
N=3; 3(3+1) iS a multiple of 3

Now go just a little further....
N=4; 4(4+1) is NOT a multiple of 3
N=5; 5(5+1) IS a multiple of 3
N=6; 6(6+1) IS a multiple of 3

Now we can see the sequence: NOT, IS, IS, NOT, IS, IS

So, 2 out of every 3 numbers in the sequence will give you a multiple of 3. This pattern repeats. Working up to 99, we would have 33 "sets" of NOT/IS/IS.

It will clearly be 2/3.

Final Answer: D

Keep this logic/approach in mind, especially if you don't immediately see the pattern behind a question. There's NO time to stare on Test Day; you're better served brute-forcing your way through this question and discovering the pattern than hoping that it will just "come to you."

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