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An integer \(\mathrm{n}\) between \(1\) and \(99\), inclusive, is to be chosen at random. What is the probability that \(\mathrm{n(n+1)}\) will be divisible by \(3\)?
Let's break down what we're being asked to find. We have integers from 1 to 99, and we want to know: if we pick one at random, what's the chance that when we multiply it by the next integer, the result is divisible by 3?
In everyday terms: if n is our chosen number, then \(\mathrm{n(n+1)}\) means we multiply n by the number that comes right after it. For example, if \(\mathrm{n = 5}\), then \(\mathrm{n(n+1) = 5 \times 6 = 30}\).
We need to count how many values of n from 1 to 99 make \(\mathrm{n(n+1)}\) divisible by 3, then divide by 99 (the total number of possibilities).
Process Skill: TRANSLATE - Converting the probability question into a counting problem
Here's a key insight: when we divide any integer by 3, we can only get three possible remainders: 0, 1, or 2.
Let's think about this with examples:
So every integer n falls into exactly one of these three categories. Now, for \(\mathrm{n(n+1)}\) to be divisible by 3, either n or \(\mathrm{(n+1)}\) needs to be divisible by 3.
Why? Because if a product is divisible by 3, then at least one of its factors must be divisible by 3.
Let's examine what happens in each case:
Case 1: n gives remainder 0 when divided by 3
This means n is divisible by 3. Examples: \(\mathrm{n = 3, 6, 9, 12...}\)
Since n is already divisible by 3, then \(\mathrm{n(n+1)}\) will definitely be divisible by 3.
✓ This case works!
Case 2: n gives remainder 1 when divided by 3
This means \(\mathrm{n = 3k + 1}\) for some integer k. Examples: \(\mathrm{n = 1, 4, 7, 10...}\)
Then \(\mathrm{n + 1 = 3k + 2}\), which gives remainder 2 when divided by 3.
So neither n nor \(\mathrm{(n+1)}\) is divisible by 3, which means \(\mathrm{n(n+1)}\) is not divisible by 3.
✗ This case doesn't work.
Case 3: n gives remainder 2 when divided by 3
This means \(\mathrm{n = 3k + 2}\) for some integer k. Examples: \(\mathrm{n = 2, 5, 8, 11...}\)
Then \(\mathrm{n + 1 = 3k + 3 = 3(k + 1)}\), which is divisible by 3!
Since \(\mathrm{(n+1)}\) is divisible by 3, then \(\mathrm{n(n+1)}\) will be divisible by 3.
✓ This case works!
Process Skill: CONSIDER ALL CASES - Systematically checking each possible remainder
From our analysis above:
So 2 out of 3 cases work!
Since integers are evenly distributed among these three remainder classes, and we want the probability for a randomly chosen integer, the answer is:
Probability = \(\frac{\text{Number of favorable cases}}{\text{Total number of cases}} = \frac{2}{3}\)
Let's verify with a few concrete examples:
This confirms our pattern: the probability is \(\frac{2}{3}\).
The probability that \(\mathrm{n(n+1)}\) will be divisible by 3 is \(\frac{2}{3}\), which corresponds to answer choice D.
1. Misunderstanding the divisibility requirement
Students may think that BOTH n AND \(\mathrm{(n+1)}\) need to be divisible by 3, rather than understanding that only ONE of them needs to be divisible by 3 for their product to be divisible by 3. This fundamental misunderstanding of divisibility rules will lead to incorrect case analysis.
2. Attempting to count specific numbers instead of using patterns
Some students may try to manually check each number from 1 to 99 (like testing \(1\times2=2\), \(2\times3=6\), \(3\times4=12\), etc.) rather than recognizing that they can use remainder patterns when dividing by 3. This approach is time-consuming and prone to counting errors.
3. Forgetting that consecutive integers have a special relationship with divisibility
Students might not realize that among any two consecutive integers n and \(\mathrm{(n+1)}\), exactly one of them will either be divisible by 3 or leave a remainder pattern that makes their product divisible by 3. Missing this insight leads to a much more complicated solution approach.
1. Incorrectly analyzing the remainder cases
When checking Case 2 (n gives remainder 1), students might incorrectly conclude that \(\mathrm{n(n+1)}\) is divisible by 3. For example, if \(\mathrm{n = 4}\) (remainder 1), then \(\mathrm{n+1 = 5}\) (remainder 2), and \(4\times5 = 20\), which is NOT divisible by 3. Students may make arithmetic errors in checking these cases.
2. Mixing up which cases work and which don't
Students may correctly identify the three remainder cases (0, 1, 2) but then incorrectly determine which ones result in \(\mathrm{n(n+1)}\) being divisible by 3. They might conclude that only Case 1 works, or that all three cases work, leading to answers like \(\frac{1}{3}\) or 1 respectively.
3. Making errors in the consecutive integer analysis
When analyzing \(\mathrm{n+1}\) in each case, students might incorrectly calculate the remainder. For instance, if n gives remainder 2 (so \(\mathrm{n = 3k+2}\)), they might incorrectly think \(\mathrm{n+1 = 3k+3}\) gives remainder 0 instead of recognizing that \(\mathrm{3k+3 = 3(k+1)}\) is actually divisible by 3.
1. Calculating the probability incorrectly
Even if students correctly identify that 2 out of 3 remainder cases work, they might calculate the final probability incorrectly. They may think the answer is \(\frac{2}{99}\) (2 favorable outcomes out of 99 total) rather than understanding that it's \(\frac{2}{3}\) (2 favorable remainder classes out of 3 total remainder classes).
2. Selecting the complementary probability
Students who correctly determine that \(\frac{2}{3}\) of cases work might accidentally select \(\frac{1}{3}\) (answer choice B) thinking about the cases that DON'T work, rather than selecting \(\frac{2}{3}\) (answer choice D) for the cases that DO work.