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An integer n between 1 and 99, inclusive, is to be chosen at 26 Mar 2021, 07:15
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Bunuel DensetsuNo VeritasKarishma I have a question about applying a method from another question to this question here (https://gmatclub.com/forum/if-an-intege ... 22078.html)

DensetsuNo gave the following quick solution for the question below:
Quote:
3-sec solution:


If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

Sequence of 4 numbers: 4
Numbers we have: 2

Answer: 1-2/4 = 1/2

You can see how this method works also on harder questions letting us solve them in a few seconds.
eg. If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

Sequence of 8 numbers: 8
Numbers we have: 3

Answer 1-3/8= 5/8.
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I applied the same methodology to this question and got 1-2/3 = 1/3 as the answer. Do you mind telling me where I got wrong in the application of the method as I am getting confused here. I do understand why 2/3 is the correct answer, but just want to know what is wrong with the method outlined above too.

Many thanks!
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An integer n between 1 and 99, inclusive, is to be chosen at 30 Mar 2021, 00:02
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salopian
Bunuel DensetsuNo VeritasKarishma I have a question about applying a method from another question to this question here (https://gmatclub.com/forum/if-an-intege ... 22078.html)

DensetsuNo gave the following quick solution for the question below:
Quote:
3-sec solution:


If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4?

Sequence of 4 numbers: 4
Numbers we have: 2

Answer: 1-2/4 = 1/2

You can see how this method works also on harder questions letting us solve them in a few seconds.
eg. If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

Sequence of 8 numbers: 8
Numbers we have: 3

Answer 1-3/8= 5/8.

I applied the same methodology to this question and got 1-2/3 = 1/3 as the answer. Do you mind telling me where I got wrong in the application of the method as I am getting confused here. I do understand why 2/3 is the correct answer, but just want to know what is wrong with the method outlined above too.

Many thanks!
Show more

All such "methods" come with a long list of constraints which must be satisfied to apply them making them pretty much useless.

Use logic:
You have 99 integers: 1-2-3... 4-5-6 ... 7-8-9 ... 97-98-99
They form perfect sets of 3.
If you take n = 1, n(n+1) will not be divisible by 3.
If you take n = 2, n(n+1) will be divisible by 3.
If you take n = 3, n(n+1) will be divisible by 3.

Same logic will apply to each of the sets of 3 numbers. For exactly 2 numbers, the product will be divisible by 3. That is why you get the answer 2/3.

Note that you could be given a range of 1-100 numbers.
Then you have 33 sets of 3 numbers each and an extra 100 at the end which if picked will not lead to a product divisible by 3.
So out of 100 numbers, there would be 34 numbers which if picked, will not lead to a product divisible by 3.
In that case, probability of a product divisible by 3 will be 66/100.
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An integer n between 1 and 99, inclusive, is to be chosen at 31 Jan 2025, 02:00
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We have to pick n from 1 to 99 inclusive, so the total number of sets of n(n+1) divisible by 3 should be 65, and not 66.

If we have to make such sets from 1 to 10, we get these sets {1,2}, {2,3}, {3,4}, {4,5}, {5,6}, {6,7}, {7,8}, {8,9}, {9,10}. Out of these - {2,3}, {3,4}, {5,6}, {6,7}, {8,9}, {9,10} are divisible by 3. So 6 out of 9.

But we need to stop at 99, so {99,100} will not be part of the list, so shouldn't the number of sets be 65 and not 66.

Bunuel ScottTargetTestPrep GMATBusters Please clarify.
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An integer n between 1 and 99, inclusive, is to be chosen at 31 Jan 2025, 02:28
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GGGMAT2
We have to pick n from 1 to 99 inclusive, so the total number of sets of n(n+1) divisible by 3 should be 65, and not 66.

If we have to make such sets from 1 to 10, we get these sets {1,2}, {2,3}, {3,4}, {4,5}, {5,6}, {6,7}, {7,8}, {8,9}, {9,10}. Out of these - {2,3}, {3,4}, {5,6}, {6,7}, {8,9}, {9,10} are divisible by 3. So 6 out of 9.

But we need to stop at 99, so {99,100} will not be part of the list, so shouldn't the number of sets be 65 and not 66.

Bunuel ScottTargetTestPrep GMATBusters Please clarify.
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{99, 100} => {n, n+1} will be a part since you have to pick n from 1 to 99, and not n+1.
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An integer n between 1 and 99, inclusive, is to be chosen at 31 Jan 2025, 02:40
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Thanks, that helps!

Krunaal
GGGMAT2
We have to pick n from 1 to 99 inclusive, so the total number of sets of n(n+1) divisible by 3 should be 65, and not 66.

If we have to make such sets from 1 to 10, we get these sets {1,2}, {2,3}, {3,4}, {4,5}, {5,6}, {6,7}, {7,8}, {8,9}, {9,10}. Out of these - {2,3}, {3,4}, {5,6}, {6,7}, {8,9}, {9,10} are divisible by 3. So 6 out of 9.

But we need to stop at 99, so {99,100} will not be part of the list, so shouldn't the number of sets be 65 and not 66.

Bunuel ScottTargetTestPrep GMATBusters Please clarify.
{99, 100} => {n, n+1} will be a part since you have to pick n from 1 to 99, and not n+1.
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An integer n between 1 and 99, inclusive, is to be chosen at 20 Apr 2025, 11:00
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My solution:-
take first 9 digits
out of this when n= 3 or 6 or 9, the given expression will be divisible by 3
Then, when n=2 0r 5 or 8, the given expression will be division by 3
Total= 6 out of 9 first digits are divisible by 3
so 6/9= 2/3 is the answer
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An integer n between 1 and 99, inclusive, is to be chosen at 22 Sep 2025, 23:03
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Hey there! This is a classic divisibility question that trips up a lot of students, but once you see the pattern, it becomes quite manageable.

Let's think about this step by step. We need to find when \(n(n+1)\) is divisible by 3, where n is any integer from 1 to 99.

Here's the key insight: For a product to be divisible by 3, at least one of its factors must be divisible by 3. So either n or (n+1) needs to be divisible by 3.

Step 1: Think in terms of remainders
When you divide any integer by 3, you can only get three possible remainders: 0, 1, or 2. Every integer falls into exactly one of these categories.

Step 2: Test each remainder case

Case 1: n gives remainder 0 when divided by 3
This means n is divisible by 3 (like n = 3, 6, 9, 12...)
Since n is already divisible by 3, then \(n(n+1)\) will definitely be divisible by 3.
✓ This case works!

Case 2: n gives remainder 1 when divided by 3
Examples: n = 1, 4, 7, 10...
Then n + 1 gives remainder 2 when divided by 3.
So neither n nor (n+1) is divisible by 3, which means \(n(n+1)\) is not divisible by 3.
✗ This case doesn't work.

Case 3: n gives remainder 2 when divided by 3
Examples: n = 2, 5, 8, 11...
Then n + 1 gives remainder 0 when divided by 3, so (n+1) is divisible by 3!
Since (n+1) is divisible by 3, then \(n(n+1)\) will be divisible by 3.
✓ This case works!

Step 3: Count the favorable outcomes
From our analysis: Cases 1 and 3 work, but Case 2 doesn't.
So 2 out of 3 cases work.

Since integers are evenly distributed among these three remainder classes, the probability is:
\(\frac{2}{3}\)

Answer: D) 2/3

You can verify this with examples: n=1 gives 1×2=2 (not divisible by 3), n=2 gives 2×3=6 (divisible by 3), n=3 gives 3×4=12 (divisible by 3), and so on.

For the complete systematic framework that applies to all consecutive integer divisibility problems, along with alternative solution methods and common trap patterns to avoid, you can check out the detailed explanation on Neuron. You can also practice with comprehensive solutions for other official questions on Neuron to build pattern recognition for similar probability and divisibility questions.
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An integer n between 1 and 99, inclusive, is to be chosen at 16 Nov 2025, 01:18
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Can you please help with the additional step in the 2nd case?
Raihanuddin
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
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An integer n between 1 and 99, inclusive, is to be chosen at 17 Nov 2025, 02:37
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dheeraj24
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 :)
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Yes, it seems right according to me.
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