When x is divided by 2, remainder is 1 and y is divided by 8, remainder is 2. Which of...

1. Translate the problem requirements: We need to express x and y in terms of their division conditions (x leaves remainder 1 when divided by 2, y leaves remainder 2 when divided by 8), then find which answer choice could be a value of 2x + y.
2. Express variables using remainder conditions: Write x and y in their general forms based on the given remainder conditions to capture all possible values.
3. Substitute into the target expression: Replace x and y in the expression 2x + y with their general forms to see what values are possible.
4. Test specific values against answer choices: Use simple integer values to generate concrete examples of 2x + y and match them with the given options.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what the problem is telling us in plain English:

• When x is divided by 2, the remainder is 1. This means x could be numbers like 1, 3, 5, 7, 9, 11, etc. - basically any odd number.
• When y is divided by 8, the remainder is 2. This means y could be numbers like 2, 10, 18, 26, 34, etc. - numbers that are 2 more than a multiple of 8.

We need to find which of the given answer choices could be a value of \(\mathrm{2x + y}\).

Process Skill: TRANSLATE - Converting the remainder conditions into understandable patterns

2. Express variables using remainder conditions

Now let's write this more systematically:

Since x leaves remainder 1 when divided by 2, we can say x is any odd number. In mathematical terms: \(\mathrm{x = 2k + 1}\), where k is any integer (0, 1, 2, 3, ...).

Since y leaves remainder 2 when divided by 8, we can write: \(\mathrm{y = 8m + 2}\), where m is any integer (0, 1, 2, 3, ...).

Let's check with examples:
• For x: when \(\mathrm{k = 0}\), \(\mathrm{x = 1}\); when \(\mathrm{k = 1}\), \(\mathrm{x = 3}\); when \(\mathrm{k = 2}\), \(\mathrm{x = 5}\), etc.
• For y: when \(\mathrm{m = 0}\), \(\mathrm{y = 2}\); when \(\mathrm{m = 1}\), \(\mathrm{y = 10}\); when \(\mathrm{m = 2}\), \(\mathrm{y = 18}\), etc.

3. Substitute into the target expression

Now let's find what \(\mathrm{2x + y}\) equals using our expressions:

\(\mathrm{2x + y = 2(2k + 1) + (8m + 2)}\)
\(= 4k + 2 + 8m + 2\)
\(= 4k + 8m + 4\)
\(= 4(k + 2m + 1)\)

This tells us that \(\mathrm{2x + y}\) must always be a multiple of 4. Let's use this insight to check our answer choices.

4. Test specific values against answer choices

Let's try some simple values and see what we get:

When \(\mathrm{k = 0}\) and \(\mathrm{m = 0}\): \(\mathrm{x = 1, y = 2}\)
\(\mathrm{2x + y = 2(1) + 2 = 4}\)

When \(\mathrm{k = 1}\) and \(\mathrm{m = 0}\): \(\mathrm{x = 3, y = 2}\)
\(\mathrm{2x + y = 2(3) + 2 = 8}\)

When \(\mathrm{k = 0}\) and \(\mathrm{m = 1}\): \(\mathrm{x = 1, y = 10}\)
\(\mathrm{2x + y = 2(1) + 10 = 12}\)

When \(\mathrm{k = 1}\) and \(\mathrm{m = 1}\): \(\mathrm{x = 3, y = 10}\)
\(\mathrm{2x + y = 2(3) + 10 = 16}\)

Process Skill: APPLY CONSTRAINTS - Using the multiple of 4 pattern to eliminate wrong answers

Looking at our answer choices:

1. 10 - not a multiple of 4
2. 11 - not a multiple of 4
3. 12 - IS a multiple of 4, and we found this value above!
4. 13 - not a multiple of 4
5. 14 - not a multiple of 4

4. Final Answer

The answer is C) 12.

We verified this works with \(\mathrm{x = 1}\) and \(\mathrm{y = 10}\):
• \(\mathrm{x = 1}\) leaves remainder 1 when divided by 2 ✓
• \(\mathrm{y = 10}\) leaves remainder 2 when divided by 8 \(\mathrm{(10 = 8 + 2)}\) ✓
• \(\mathrm{2x + y = 2(1) + 10 = 12}\) ✓

Additionally, since \(\mathrm{2x + y}\) must always be a multiple of 4, only choice C satisfies this requirement.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding remainder notation: Students often confuse "x divided by 2 has remainder 1" with "\(\mathrm{x = 1/2}\)" or think it means x must equal 1. They fail to recognize that this describes a pattern where x could be 1, 3, 5, 7, etc. (any odd number).

2. Not recognizing the constraint pattern: Students may attempt to solve by randomly plugging in values from the answer choices without first establishing the mathematical relationship that \(\mathrm{x = 2k + 1}\) and \(\mathrm{y = 8m + 2}\). This leads to inefficient trial-and-error rather than systematic analysis.

3. Missing the key insight about multiples: Students often fail to recognize that they should substitute the expressions for x and y into \(\mathrm{2x + y}\) to find that the result must always be a multiple of 4. This insight dramatically simplifies the problem by eliminating most answer choices immediately.

Errors while executing the approach

1. Algebraic manipulation errors: When expanding \(\mathrm{2x + y = 2(2k + 1) + (8m + 2)}\), students commonly make errors like forgetting to distribute the 2, getting \(\mathrm{2k + 1 + 8m + 2}\) instead of \(\mathrm{4k + 2 + 8m + 2}\), leading to incorrect final expressions.

2. Remainder verification mistakes: When checking specific values, students may incorrectly verify remainders. For example, when checking \(\mathrm{y = 10}\) divided by 8, they might miscalculate and think the remainder is 1 instead of 2, or forget that \(\mathrm{10 = 8(1) + 2}\).

3. Insufficient value testing: Students might test only one combination of k and m values, missing the pattern that emerges. They may get lucky with one calculation but fail to verify their understanding with additional examples.

Errors while selecting the answer

1. Ignoring the multiple of 4 constraint: Even after correctly determining that \(\mathrm{2x + y}\) must be a multiple of 4, students might still consider non-multiples of 4 from the answer choices, essentially abandoning their mathematical insight in favor of random guessing.

2. Calculation verification errors: Students may correctly identify that the answer should be 12, but then second-guess themselves and fail to properly verify that \(\mathrm{x = 1, y = 10}\) satisfies both original remainder conditions, leading them to change their answer unnecessarily.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose smart values for x
Since x divided by 2 gives remainder 1, x must be odd. Let's choose \(\mathrm{x = 1}\) (the smallest positive odd integer).
Check: \(\mathrm{1 ÷ 2 = 0}\) remainder 1 ✓

Step 2: Choose smart values for y
Since y divided by 8 gives remainder 2, let's choose \(\mathrm{y = 2}\) (the smallest positive integer with this property).
Check: \(\mathrm{2 ÷ 8 = 0}\) remainder 2 ✓

Step 3: Calculate 2x + y with our chosen values
\(\mathrm{2x + y = 2(1) + 2 = 2 + 2 = 4}\)
This value (4) is not among the answer choices, so let's try other valid values.

Step 4: Try the next valid value for x
Next odd number: \(\mathrm{x = 3}\)
Check: \(\mathrm{3 ÷ 2 = 1}\) remainder 1 ✓
Keep \(\mathrm{y = 2}\)
\(\mathrm{2x + y = 2(3) + 2 = 6 + 2 = 8}\)
Still not in the answer choices.

Step 5: Try x = 5
\(\mathrm{x = 5: 5 ÷ 2 = 2}\) remainder 1 ✓
\(\mathrm{y = 2}\)
\(\mathrm{2x + y = 2(5) + 2 = 10 + 2 = 12}\)
This matches answer choice C!

Step 6: Verify with another valid y value
Next valid y: \(\mathrm{y = 10}\) (since \(\mathrm{10 ÷ 8 = 1}\) remainder 2)
Using \(\mathrm{x = 1: 2(1) + 10 = 12}\)
This also gives us 12, confirming our answer.

Conclusion: The value 12 can be achieved with multiple combinations of valid x and y values, making C the correct answer.