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If an integer \(\mathrm{n}\) is to be selected at random from \(\mathrm{1}\) to \(\mathrm{100}\), inclusive, what is probability \(\mathrm{n(n+1)}\) will be divisible by \(\mathrm{4}\)?
Let's start by understanding what we're looking for in plain English. We have integers from 1 to 100, 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 4?
For example:
So we need to count how many values of n from 1 to 100 make \(\mathrm{n(n+1)}\) divisible by 4.
Process Skill: TRANSLATE - Converting the probability question into a counting problem
Here's the key insight: when we multiply two consecutive integers, we're guaranteed that one of them is even (since consecutive numbers alternate between odd and even). But for the product to be divisible by 4, we need the product to have at least two factors of 2.
Let's think about this systematically by looking at what happens when we divide n by 4. Any integer n can be written as one of these forms:
Let's check each case with concrete examples:
Case 1: n leaves remainder 0 when divided by 4 (n is divisible by 4)
Example: \(\mathrm{n = 4}\), so \(\mathrm{n(n+1) = 4 \times 5 = 20}\)
Since n is already divisible by 4, the product \(\mathrm{n(n+1)}\) is automatically divisible by 4.
Case 2: n leaves remainder 1 when divided by 4
Example: \(\mathrm{n = 5}\), so \(\mathrm{n(n+1) = 5 \times 6 = 30}\)
Here n is odd, but \(\mathrm{n+1 = 6}\) is divisible by 2 but not by 4. So the product has only one factor of 2, not enough for divisibility by 4.
Case 3: n leaves remainder 2 when divided by 4
Example: \(\mathrm{n = 6}\), so \(\mathrm{n(n+1) = 6 \times 7 = 42}\)
Here \(\mathrm{n = 6}\) is divisible by 2 but not by 4, and \(\mathrm{n+1 = 7}\) is odd. Again, only one factor of 2 in the product.
Case 4: n leaves remainder 3 when divided by 4
Example: \(\mathrm{n = 7}\), so \(\mathrm{n(n+1) = 7 \times 8 = 56}\)
Here n is odd, but \(\mathrm{n+1 = 8}\) is divisible by 4. So the product is divisible by 4.
Process Skill: CONSIDER ALL CASES - Systematically checking all possible remainders
From our analysis, \(\mathrm{n(n+1)}\) is divisible by 4 when:
Let's verify this pattern with more examples:
And checking the cases that shouldn't work:
Our pattern holds!
Now we need to count how many integers from 1 to 100 fall into our favorable cases.
Case 1: \(\mathrm{n \equiv 0 \pmod{4}}\)
These are: 4, 8, 12, 16, ..., 100
This is an arithmetic sequence with first term 4, common difference 4, and last term 100.
Number of terms = \(\mathrm{\frac{(100 - 4)}{4} + 1 = \frac{96}{4} + 1 = 24 + 1 = 25}\)
Case 2: \(\mathrm{n \equiv 3 \pmod{4}}\)
These are: 3, 7, 11, 15, ..., 99
This is an arithmetic sequence with first term 3, common difference 4, and last term 99.
Number of terms = \(\mathrm{\frac{(99 - 3)}{4} + 1 = \frac{96}{4} + 1 = 24 + 1 = 25}\)
Total favorable outcomes: 25 + 25 = 50
Total possible outcomes: 100
Probability: \(\mathrm{\frac{50}{100} = \frac{1}{2}}\)
Process Skill: SIMPLIFY - Reducing the fraction to its simplest form
The probability that \(\mathrm{n(n+1)}\) will be divisible by 4 is \(\mathrm{\frac{1}{2}}\).
This matches answer choice C.
To verify our logic: exactly half of all integers from 1 to 100 will produce products divisible by 4 when multiplied by their successor. This makes intuitive sense because the favorable cases (remainders 0 and 3 when divided by 4) represent exactly 2 out of every 4 consecutive integers.
1. Misunderstanding what "divisible by 4" means for products
Students often think that if either n or (n+1) is divisible by 4, then their product is divisible by 4. This leads them to only look for multiples of 4, missing the crucial insight that we need at least two factors of 2 in the product. They fail to recognize that when \(\mathrm{n \equiv 3 \pmod{4}}\), then \(\mathrm{n+1 \equiv 0 \pmod{4}}\), making the product divisible by 4.
2. Attempting to check every case individually instead of finding a pattern
Rather than systematically analyzing the four possible remainders when n is divided by 4, students may try to check each number from 1 to 100 individually. This approach is time-consuming and prone to computational errors, especially under test conditions where time management is crucial.
3. Overlooking the consecutive integer property
Students may forget that n and (n+1) are consecutive integers, which guarantees that exactly one of them is even. This fundamental property is key to understanding why only certain remainders work, and missing it leads to incorrect analysis of when the product has sufficient factors of 2.
1. Incorrectly counting arithmetic sequences
When counting integers of the form 4k and 4k+3 from 1 to 100, students often make mistakes in the arithmetic sequence formula. Common errors include forgetting to add 1 when using the formula (last term - first term)/common difference, or miscalculating the last term in each sequence (using 96 instead of 99 for the sequence 3, 7, 11, ...).
2. Computational errors when checking the pattern
During verification, students may make simple multiplication or division errors. For example, when checking if \(\mathrm{7 \times 8 = 56}\) is divisible by 4, they might incorrectly calculate \(\mathrm{56 \div 4 = 13}\) instead of 14, leading them to doubt their pattern and potentially abandon the correct approach.
3. Mixing up the modular arithmetic cases
Students may correctly identify that they need to check remainders 0, 1, 2, and 3, but then incorrectly determine which cases work. They might conclude that remainders 1 and 2 are favorable instead of 0 and 3, especially if they make errors in their test calculations.
No likely faltering points - Once students correctly count 50 favorable outcomes out of 100 total outcomes, the probability calculation \(\mathrm{\frac{50}{100} = \frac{1}{2}}\) is straightforward, and \(\mathrm{\frac{1}{2}}\) directly matches answer choice C without requiring further manipulation.