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If \(\mathrm{x}\) is an integer and \(\mathrm{y = 3x + 2}\), which of the following CANNOT be a divisor of \(\mathrm{y}\)?
Let's break down what we're looking for in plain English. We have a formula \(\mathrm{y = 3x + 2}\), where x can be any integer (positive, negative, or zero). We need to find which number from our answer choices can NEVER divide y evenly - meaning when we divide y by that number, we never get a whole number result with no remainder.
A number is called a "divisor" of y when it divides y exactly with no remainder left over. So we're hunting for the one answer choice that will ALWAYS leave a remainder when we try to divide any possible y-value by it.
Process Skill: TRANSLATE
Let's substitute some consecutive integer values for x and see what y-values we get. This will help us spot patterns:
When \(\mathrm{x = 0: y = 3(0) + 2 = 2}\)
When \(\mathrm{x = 1: y = 3(1) + 2 = 5}\)
When \(\mathrm{x = 2: y = 3(2) + 2 = 8}\)
When \(\mathrm{x = 3: y = 3(3) + 2 = 11}\)
When \(\mathrm{x = 4: y = 3(4) + 2 = 14}\)
When \(\mathrm{x = 5: y = 3(5) + 2 = 17}\)
When \(\mathrm{x = 6: y = 3(6) + 2 = 20}\)
When \(\mathrm{x = 7: y = 3(7) + 2 = 23}\)
When \(\mathrm{x = 8: y = 3(8) + 2 = 26}\)
So our sequence of possible y-values starts: 2, 5, 8, 11, 14, 17, 20, 23, 26, ...
Process Skill: SIMPLIFY
Now let's check each answer choice to see if any of our y-values can be divided evenly by them:
Choice (A): Can 4 be a divisor?
Looking at our y-values: 2, 5, 8, 11, 14, 17, 20, 23, 26...
\(\mathrm{8 ÷ 4 = 2}\) exactly, and \(\mathrm{20 ÷ 4 = 5}\) exactly
So 4 CAN be a divisor of y.
Choice (B): Can 5 be a divisor?
\(\mathrm{5 ÷ 5 = 1}\) exactly, and \(\mathrm{20 ÷ 5 = 4}\) exactly
So 5 CAN be a divisor of y.
Choice (D): Can 7 be a divisor?
When \(\mathrm{x = 12: y = 3(12) + 2 = 38}\), and \(\mathrm{38 = 7 × 5 + 3}\)
Let me try \(\mathrm{x = 19: y = 3(19) + 2 = 59}\), and \(\mathrm{59 = 7 × 8 + 3}\)
Let me try \(\mathrm{x = 5: y = 3(5) + 2 = 17}\), and \(\mathrm{17 = 7 × 2 + 3}\)
Actually, let me try \(\mathrm{x = 16: y = 3(16) + 2 = 50}\), and \(\mathrm{50 = 7 × 7 + 1}\)
Trying \(\mathrm{x = 17: y = 3(17) + 2 = 53}\), and \(\mathrm{53 = 7 × 7 + 4}\)
Trying \(\mathrm{x = 18: y = 3(18) + 2 = 56}\), and \(\mathrm{56 ÷ 7 = 8}\) exactly
So 7 CAN be a divisor of y.
Choice (E): Can 8 be a divisor?
\(\mathrm{8 ÷ 8 = 1}\) exactly, and 24 would work (when x = 22/3... wait, x must be an integer)
Let me check: when \(\mathrm{x = 14, y = 3(14) + 2 = 44}\), and \(\mathrm{44 = 8 × 5 + 4}\)
When \(\mathrm{x = 2, y = 8}\), and \(\mathrm{8 ÷ 8 = 1}\) exactly
So 8 CAN be a divisor of y.
Choice (C): Can 6 be a divisor?
Let's check our y-values systematically:
\(\mathrm{2 ÷ 6 = 0}\) remainder 2
\(\mathrm{5 ÷ 6 = 0}\) remainder 5
\(\mathrm{8 ÷ 6 = 1}\) remainder 2
\(\mathrm{11 ÷ 6 = 1}\) remainder 5
\(\mathrm{14 ÷ 6 = 2}\) remainder 2
\(\mathrm{17 ÷ 6 = 2}\) remainder 5
\(\mathrm{20 ÷ 6 = 3}\) remainder 2
\(\mathrm{23 ÷ 6 = 3}\) remainder 5
\(\mathrm{26 ÷ 6 = 4}\) remainder 2
Notice a pattern: when we divide y by 6, we always get remainder 2 or remainder 5, but never remainder 0!
Process Skill: INFER
From our systematic testing, we can see that:
We can verify this pattern: since \(\mathrm{y = 3x + 2}\), and 3x is always a multiple of 3, we have y = (multiple of 3) + 2. When we divide any multiple of 3 by 6, we get remainder 0 or 3. Adding 2 to this gives us remainder 2 or 5, but never 0.
The answer is (C) 6. Through systematic testing of y-values, we discovered that when \(\mathrm{y = 3x + 2}\) is divided by 6, the remainder is always either 2 or 5, never 0. Since exact divisibility requires remainder 0, the number 6 cannot be a divisor of y for any integer value of x.
1. Misinterpreting "CANNOT be a divisor"
Students often miss the word "CANNOT" and look for numbers that CAN divide y, leading them to select an answer choice that actually works as a divisor. This is a critical reading error that completely reverses the solution approach.
2. Forgetting that x must be an integer
Students may overlook the constraint that "x is an integer" and consider fractional values of x in their analysis. This constraint is crucial because it limits the possible values of y to a specific arithmetic sequence.
3. Testing only positive values of x
Some students forget that integers include negative numbers and zero, testing only positive values like x = 1, 2, 3... This incomplete approach might miss important patterns or lead to incorrect conclusions about which numbers can be divisors.
1. Arithmetic errors in calculating y-values
When substituting values like x = 3, 4, 5 into \(\mathrm{y = 3x + 2}\), students may make simple calculation mistakes (e.g., calculating \(\mathrm{3(4) + 2 = 16}\) instead of 14), which leads to wrong y-values and incorrect divisibility testing.
2. Incorrect division and remainder calculations
When checking if numbers like 6 divide y-values, students may incorrectly calculate remainders. For example, when dividing 14 by 6, they might say the remainder is 3 instead of 2, missing the crucial pattern that remainders are always 2 or 5.
3. Stopping testing too early
Students might test only one or two values and conclude a number can be a divisor without checking enough cases. For instance, finding that 8 doesn't divide the first few y-values and concluding 8 can never be a divisor, without testing x = 2 where y = 8.
1. Selecting a number that CAN be a divisor
After correctly identifying patterns, students may accidentally select an answer choice that actually works as a divisor (like 4 or 5) instead of the one that cannot be a divisor (6), due to the reverse logic required by "CANNOT."
2. Second-guessing the pattern recognition
Even after correctly observing that dividing y by 6 always gives remainder 2 or 5 (never 0), students might doubt their pattern and select a different answer, thinking they made an error in their systematic approach.
Step 1: Choose strategic values of x
To find patterns in divisibility, let's test x = 0, 1, 2, 3, 4, 5 to generate a representative set of y-values and observe remainder patterns.
Step 2: Calculate corresponding y-values
• When \(\mathrm{x = 0: y = 3(0) + 2 = 2}\)
• When \(\mathrm{x = 1: y = 3(1) + 2 = 5}\)
• When \(\mathrm{x = 2: y = 3(2) + 2 = 8}\)
• When \(\mathrm{x = 3: y = 3(3) + 2 = 11}\)
• When \(\mathrm{x = 4: y = 3(4) + 2 = 14}\)
• When \(\mathrm{x = 5: y = 3(5) + 2 = 17}\)
Step 3: Test divisibility by each answer choice
Testing (A) 4: \(\mathrm{y = 8 ÷ 4 = 2}\) exactly ✓ (4 can be a divisor)
Testing (B) 5: \(\mathrm{y = 5 ÷ 5 = 1}\) exactly ✓ (5 can be a divisor)
Testing (C) 6: Check all y-values:
• \(\mathrm{2 ÷ 6}\) = remainder 2
• \(\mathrm{5 ÷ 6}\) = remainder 5
• \(\mathrm{8 ÷ 6}\) = remainder 2
• \(\mathrm{11 ÷ 6}\) = remainder 5
• \(\mathrm{14 ÷ 6}\) = remainder 2
• \(\mathrm{17 ÷ 6}\) = remainder 5
Pattern: remainders alternate between 2 and 5, never 0
Testing (D) 7: \(\mathrm{y = 2 ÷ 7}\) = remainder 2, but we need to check if 7 ever divides y exactly
When \(\mathrm{x = 4: y = 14}\), and \(\mathrm{14 ÷ 7 = 2}\) exactly ✓ (7 can be a divisor)
Testing (E) 8: \(\mathrm{y = 8 ÷ 8 = 1}\) exactly ✓ (8 can be a divisor)
Step 4: Identify the pattern for 6
The key insight from our smart numbers: when \(\mathrm{y = 3x + 2}\) is divided by 6, we only get remainders of 2 (when x is even) or 5 (when x is odd). We never get remainder 0, which means 6 can never be a divisor of y.
Answer: (C) 6