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How many positive integers less than \(\mathrm{50}\) have a reminder \(\mathrm{1}\) when divided by \(\mathrm{3}\)?
Let's break down what this problem is asking in plain English. We want to find positive integers (counting numbers like 1, 2, 3, etc.) that are less than 50, and when we divide them by 3, we get a remainder of 1.
Think of it this way: if I have a number and I divide it by 3, I might get some whole number with a little bit left over. That "little bit left over" is the remainder. We want cases where that leftover piece equals exactly 1.
Mathematically, we're looking for numbers that can be written as \(3k + 1\), where \(k\) is some non-negative whole number (0, 1, 2, 3, ...), and the result is a positive number less than 50.
Process Skill: TRANSLATE - Converting the division remainder concept into a concrete mathematical form
Let's start by finding the first several numbers that give us a remainder of 1 when divided by 3. I'll check this by actually doing the division:
The pattern is clear! Our numbers are: 1, 4, 7, 10, 13, 16, 19, ...
Notice that each number is exactly 3 more than the previous one. This makes sense because if a number has remainder 1 when divided by 3, adding 3 to it won't change the remainder.
This forms an arithmetic sequence with first term \(a_1 = 1\) and common difference \(d = 3\).
Now I need to find the smallest and largest numbers in our sequence that meet our conditions.
Smallest number: We need positive integers, so our smallest number is 1.
Largest number: We need numbers less than 50. Let me find the largest number in our sequence that's still less than 50.
Following our pattern: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49
Let me check: \(49 ÷ 3 = 16\) remainder 1 ✓
The next number would be \(49 + 3 = 52\), but that's not less than 50.
So our sequence runs from 1 to 49.
Process Skill: APPLY CONSTRAINTS - Ensuring we stay within the "less than 50" boundary
Now I need to count how many numbers are in the sequence: 1, 4, 7, 10, ..., 46, 49
Since this is an arithmetic sequence, I can use the relationship between the terms. Each term can be written as:
\(\mathrm{Term} = 1 + 3(k)\), where \(k = 0, 1, 2, 3, ...\)
For the last term: \(49 = 1 + 3k\)
Solving: \(48 = 3k\), so \(k = 16\)
This means \(k\) goes from 0 to 16, which gives us 17 different values.
Let me verify by counting directly: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49
Counting these: 17 numbers total.
There are 17 positive integers less than 50 that have a remainder of 1 when divided by 3.
This matches answer choice E) 17.
Faltering Point 1: Misunderstanding the remainder requirement
Students often confuse "remainder 01" with "remainder 1" or might think they need numbers that are divisible by 3 (remainder 0). The problem states "remainder 01" which should be interpreted as remainder 1. This confusion can lead students to look for multiples of 3 instead of numbers that leave remainder 1.
Faltering Point 2: Overlooking the constraint boundaries
Students may miss the "less than 50" constraint and include 50 in their count, or they might misinterpret "positive integers" and include 0. The problem specifically asks for positive integers (1, 2, 3, ...) that are less than 50, not less than or equal to 50.
Faltering Point 3: Not recognizing the arithmetic sequence pattern
Students might try to check each number individually from 1 to 49 without recognizing that numbers with remainder 1 when divided by 3 form an arithmetic sequence (1, 4, 7, 10, ...). This leads to a time-consuming approach and potential counting errors.
Faltering Point 1: Arithmetic errors in division checks
When manually checking remainders by division, students often make calculation mistakes. For example, they might incorrectly calculate that \(22 ÷ 3\) gives remainder 2 instead of remainder 1, or make similar arithmetic errors throughout their verification process.
Faltering Point 2: Incorrect application of arithmetic sequence formula
Students may set up the wrong equation when finding the last term. They might write \(49 = 3k + 1\) but then solve incorrectly, getting \(k = 15\) instead of \(k = 16\), or they might confuse the number of terms with the value of \(k\) (forgetting that \(k\) starts from 0).
Faltering Point 3: Missing terms in the sequence
When listing out the sequence manually, students might skip numbers or stop too early. For instance, they might forget to include 49 in their list, thinking 46 is the last valid number, or they might accidentally skip a term like 31 or 37 in their enumeration.
Faltering Point 1: Off-by-one counting error
Students correctly identify the sequence but make a counting mistake. They might count 16 terms instead of 17 because they forget that when \(k\) goes from 0 to 16, that represents 17 different values (including both endpoints). This leads them to select answer choice D) 16 instead of the correct E) 17.