An integer n between 1 and 99, inclusive, is to be chosen at

Question

An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ?

A) 1/9
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Bunuel · Accepted Answer

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

In every group of three consecutive numbers—for example, {1, 2, 3}, {4, 5, 6}, {7, 8, 9}, ..., {97, 98, 99}—exactly two numbers satisfy this condition. For instance, in {1, 2, 3}, n can be 2 or 3. Therefore, the overall probability is 2/3.

Answer: D.

Raihanuddin · Answer

In such problem always follow a simple rule.
The divisor is 3. So take number from 1 to 3 as example

If n=1, n(n+1)= Not divisible by 3
If n=2, n(n+1)= yes
If n=3, n(n+1)=yes. Don't need any more example.

So, 2 out of 3 are divisible. Hence, 2/3 is the answer.

This rule is applicable if last number(99 in this case) is divisible by 3.

If you had to pick from 1-98, still you can apply the rule but be careful. There is one more additional step.

Hopefully you can find answer now in less than 30 seconds

Raihanuddin · Answer

If it is by 8, then you have to do the above plus an additional step.

if n = 97, n(n+1) not divisible by 8
if n= 98 still not divisible by 8 and
if n= 99 also not divisible by 8

Now notice that from 1 to 96, total 24 values of n is divisible by 8. We have 3 more values as shown above. So out of 99 values 24 are divisible.

Probability = 24/99 = 8/33

You may think that where I have got 96. from 1 to 99 we can't divide 99 by 8. The highest number we can divide is 96. So we have three more numbers left. That's why we have to do the additional step. We are using cycle of (1-8), (9-16) (17-24) ..........

You can see that 2 numbers from (1-8) are divisible by 8 if you consider n(n+1). Similarly 2 numbers from (9-16) and so on

How many such pairs do we have from 1-99.? 12 pairs of 8. So 2 numbers from each pair are divisible. Therefore total 2*12= 24 are divisible out of 96 numbers. And from remaining number none is divisible by 8. So, out of 99 numbers 24 are divisible.

You can do the next question similarly.

Very important

I think in official question you will not get such number as 8 by which you can't divide last number, which is 99. So, don't worry for that. The given question in the main post is an official question and I have always got that GMAT gives you last number divisible by the divisor.

Bunuel · Answer

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dheeraj24 · Answer

Hi guys,
Could anyone tell whether my approach for this problem is correct.

Multiples of 3 between 1 and 99 are 33.
Multiples of (n+1) between 1 and 99 must also be 33.
Probability that [n(n+1)/3 ] is divisible by 3 is either n is divisible by 3 (or) (n+1) is divisible by 3 .
Probablity that n is divisible by 3 is 33/99 = 1/3;
Probabilty that n+1 is divisible by 3 is 33/99 = 1/3;
Therefore the probability that n(n+1)/3 is 1/3+ 1/3 = 2/3;
Thanks in advance

LalaB · Answer

the required number = (if n is even )+ (if n is divisible by 3) - (if n is even and divisible by 3)=
=(49+33-16)/99=66/99=2/3

ajithkumar · Answer

Any three consecutive integers will exactly have one number that is a multiple of three. So the probability of 2 consecutive integers to have one multiple of 3 is 2/3 (if the total set of numbers considered is a multiple of three, it doesn't matter if the set is {1,2..6}, {1,2,..30}, or {1,2,3...99}, the probability will always be 2/3).

Ans: D

Ans: D

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

Onthemove · Answer

Hi Bunue/ Raihanuddin

Can you use the same theory for n(n+1) divisible by 8. For example take (1,2,3,4,5,6,7,8), only 7 and 8 work so prob is 2/8 = 1/4?

Also how would you do it or even n(n+1)(n-1) divisible by 8 ? Is there a general rule/trick, might help a lot to reduce time spent on these questions.

rohitkumar1 · Answer

Would appreciate some input regarding my confusion with this problem.

If n(n+1) has to be divisible by 3, there could be 3 scenarios in all:

1) n is divisible by 3 AND (n+1) is not divisibly by 3  I see the mistake here now: if n is divisible by 3, n+1 is automatically not divisible by 3

2) n is not divisible by 3 AND (n+1) is divisible by 3  Same error as above

3) n AND (n+1) are BOTH divisible by 3  We're asked to pick only one integer n so this is invalid/unnecessary

Out of 99 elements, 33 will be of the form n divisible by 3, 33 will be of the form (n+1) divisibly by 3, and 33 will be neither.

So the total probability for cases 1-3 above, per me, should work out to something like: (33/99*33/99)*3 = (1/9)*3 = 1/3.

Please tell me what part of the above logic is flawed?

Thank you.

Edit: Have retraced my steps and commented what I think my mistakes were - would still appreciate some endorsement from other members.

EMPOWERgmatRichC · Answer

Hi All,

This question is based on pattern-matching. The good news is that even if you don't immediately spot the pattern (some people can just "see it"), you can still PROVE that a pattern exists with just a little work...

We're told that N is a number from 1 to 99, inclusive. We're asked for the probability that N(N+1) is divisible by 3....

IF....
N = 1....(1)(2) = 2 is NOT divisible by 3.
N = 2....(2)(3) = 6 IS divisible by 3
N = 3....(3)(4) = 12 IS divisible by 3.

In these first 3 examples, notice how 2 of the 3 are divisible by 3....What happens when we look at the NEXT 3 numbers....

IF....
N = 4....(4)(5) = 20 is NOT divisible by 3.
N = 5....(5)(6) = 30 IS divisible by 3
N = 6....(6)(7) = 42 IS divisible by 3.

The pattern REPEATS! This means that for ever 3 consecutive values of N, 2 of the 3 will be divisible by 3. Since we're dealing with the numbers 1 to 99, inclusive, that means we'll have 33 "sets" of 3 numbers (as shown above). Thus, 2/3 of ALL the values WILL be divisible by 3...

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

EgmatQuantExpert · Answer

Hi avgroh , You have rightly calculated that there will be 33 numbers of the form n which are divisible by 3 and 33 numbers of the form n + 1 which are divisible by 3. The question asks us the probability of product of two numbers being divisible by 3. This is possible if either of them is divisible by 3 i.e. if n is divisible by 3 or n + 1 is divisible by 3. So, it's an OR event. We don't need both the events to happen for n(n+1) to be divisible by 3. Hence we will add the probabilities of both the events happening. So we can write P(n(n+1)) is divisible by 3 = \frac{33}{99} + \frac{33}{99} = \frac{66}{99} = \frac{2}{3} Hope it's clear

Regards Harsh

MartyTargetTestPrep · Answer

If either n or (n + 1) is divisible by 3, then n(n + 1) is divisible by 3.

So, we have two situations in which n(n + 1) is divisible by 3.

n is a multiple of 3.

(n + 1) is a multiple of 3. In such a case n = (a multiple of 3) - 1.

So, to find the number of integers such that n(n + 1) is a multiple of 3, all you have to do is:

Determine how many multiples of 3 appear in 1 to 96 inclusive.

Confirm that, for every multiple of three, there is also in 1 to 96 inclusive a number one less than that multiple of 3. (For instance, if the list were 3 to 96 inclusive, then for one multiple of 3, 3 itself, there would not be a number one less that than multiple of 3.) In 1 to 96 inclusive, for every multiple of 3, there is indeed a number one less than that multiple of 3.

Multiply by 2.

This method works because it accounts for every case in which n is a multiple of 3 and every case in which n is (a multiple of 3) - 1.

Since 96 is a multiple of 3, and since, for every multiple of 3 between 1 and 96 inclusive, there is an integer value 1 less than that multiple of 3, we can simply divide 96 by 3 and multiply by 2 to calculate the number of instances such that n(n + 1) is divisible by 3.

96/3 x 2 = 64

So, the probability of choosing an n such that n(n + 1) is a multiple of 3 is 64/96 = 2/3.

peterpark · Answer

An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ?

A) 1/9
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Can someone explain to me why this method does not work?

I caught the pattern of 6 possibilities that meet the question stems criteria from 1 - 10 {2,3,5,6,8,9}.
I multiplied 6 by 10 given the range described above happens 10 times 1 -10, 11 - 20, 21 - 30, 31 - 40, and etc...
This now gives me 60/100 which simplifies to 3/5.
It seems I am missing 6 cases.

Additionally, can someone explain how they got a denominator of 99 and not 100?

EMPOWERgmatRichC · Answer

Hi peterpark,

To start, we're told that N is an integer from 1 to 99, inclusive, so your fraction will be out of 99 (not 100).

Based on your work, you actually spotted a pattern, but you didn't interpret it correctly. Instead of focusing on the 'first 10 terms', try focusing on 'groups of 3'...

IF....
N = 1....(1)(2) = 2 is NOT divisible by 3.
N = 2....(2)(3) = 6 IS divisible by 3
N = 3....(3)(4) = 12 IS divisible by 3.

In these first 3 examples, notice how 2 of the 3 are divisible by 3....What happens when we look at the NEXT 3 numbers....

IF....
N = 4....(4)(5) = 20 is NOT divisible by 3.
N = 5....(5)(6) = 30 IS divisible by 3
N = 6....(6)(7) = 42 IS divisible by 3.

The pattern REPEATS! This means that for ever 3 consecutive values of N, 2 of the 3 will be divisible by 3. Since we're dealing with the numbers 1 to 99, inclusive, that means we'll have 33 "sets" of 3 numbers (as shown above). Thus, 2/3 of ALL the values WILL be divisible by 3...

GMAT assassins aren't born, they're made,
Rich

ScottTargetTestPrep · Answer

The number of multiples of 3 from 1 to 99, inclusive, is:

(99 - 3)/3 + 1 = 33

We also should see that every number that is 1 less than a multiple of 3 will also allow n(n+1) to be divisible by 3. For example, if n = 2, then n + 1 = 3, which is divisible by 3. Since there are 33 multiples of 3, there are 33 numbers that are one less than a multiple of 3. Thus, the probability that n(n+1) will be divisible by 3 is 66/99 = 2/3.

Answer: D

GMATBusters · Answer

out of any n consecutive integers, there is always a multiple of n.  
  out of any 3 consecutive numbers, there is always a multiple of 3.

What is the probability that n(n+1) will be divisible by 3?
n and (n+1) are consecutive integers, we know that out of any 3 consecutive numbers, there is always a multiple of 3.
so, the probability that n or n+1 is multiple of 3 = 2/3
Answer D

BoscoVIX · Answer

Can someone explain why a sequence of 3 numbers?

EMPOWERgmatRichC · Answer

Hi ejonuma,

Since we're dealing with consecutive integers - and we're looking for results that are DIVISBLE BY 3 - it's likely that a pattern will exist in "groups of 3." For example:

Of the numbers 1, 2, 3, 4, 5, 6..... notice how every THIRD number is divisible by 3.

This question asks us to do a bit more work than that though, but even if you aren't thinking in terms of 'groups of 3', it does not take much effort to figure out that a pattern exists...

IF....
N = 1....(1)(2) = 2 is NOT divisible by 3.
N = 2....(2)(3) = 6 IS divisible by 3
N = 3....(3)(4) = 12 IS divisible by 3.
N = 4....(4)(5) = 20 is NOT divisible by 3.
N = 5....(5)(6) = 30 IS divisible by 3
N = 6....(6)(7) = 42 IS divisible by 3.

Notice that there is a repeating pattern! For every 3 consecutive values of N, 2 of the 3 will be divisible by 3. Since we're dealing with the numbers 1 to 99, inclusive, that means we'll have 33 "sets" of 3 numbers (as shown above).

GMAT assassins aren't born, they're made,
Rich

An integer n between 1 and 99, inclusive, is to be chosen at

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An integer n between 1 and 99, inclusive, is to be chosen at

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