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If \(\mathrm{x}\) and \(\mathrm{y}\) are positive integers such that \(\mathrm{y}\) is a multiple of 5 and \(3\mathrm{x} + 4\mathrm{y} = 200\), then \(\mathrm{x}\) must be a multiple of which of the following?
Let's start by understanding what we're given and what we need to find.
We have two positive integers, x and y, with these conditions:
• y is a multiple of 5 (meaning y could be 5, 10, 15, 20, etc.)
• \(3\mathrm{x} + 4\mathrm{y} = 200\) (this is our constraint equation)
• We need to find what number x must be divisible by
Think of this like a puzzle where we have a specific relationship between x and y, and we need to figure out what pattern x must follow.
Process Skill: TRANSLATE
Let's solve for x from our equation \(3\mathrm{x} + 4\mathrm{y} = 200\).
Starting with: \(3\mathrm{x} + 4\mathrm{y} = 200\)
Subtract \(4\mathrm{y}\) from both sides: \(3\mathrm{x} = 200 - 4\mathrm{y}\)
Divide both sides by 3: \(\mathrm{x} = \frac{200 - 4\mathrm{y}}{3}\)
Now we can see that x depends on what y is. Since we know y has to be a multiple of 5, let's use that information.
Since y is a multiple of 5, we can write \(\mathrm{y} = 5\mathrm{k}\) where k is any positive integer.
Substituting this into our expression for x:
\(\mathrm{x} = \frac{200 - 4\mathrm{y}}{3}\)
\(\mathrm{x} = \frac{200 - 4(5\mathrm{k})}{3}\)
\(\mathrm{x} = \frac{200 - 20\mathrm{k}}{3}\)
Let's factor this expression:
\(\mathrm{x} = \frac{200 - 20\mathrm{k}}{3}\)
\(\mathrm{x} = \frac{20(10 - \mathrm{k})}{3}\)
For x to be a positive integer, \((10 - \mathrm{k})\) must be divisible by 3. But more importantly, let's see what this tells us about x's divisibility.
Process Skill: MANIPULATE
Looking at our expression \(\mathrm{x} = \frac{20(10 - \mathrm{k})}{3}\), we can rewrite this as:
\(\mathrm{x} = \frac{20}{3} \times (10 - \mathrm{k})\)
For x to be an integer, \((10 - \mathrm{k})\) must be divisible by 3. Let's call \(\frac{(10 - \mathrm{k})}{3} = \mathrm{m}\), where m is an integer.
Then: \(\mathrm{x} = 20\mathrm{m}\)
This means x must always be a multiple of 20!
Since \(20 = 4 \times 5 = 2^2 \times 5\), any multiple of 20 is automatically a multiple of:
• 1, 2, 4, 5, 10, and 20
Looking at our answer choices:
A) 3 - Not necessarily (20 is not divisible by 3)
B) 6 - Not necessarily (20 is not divisible by 6)
C) 7 - Not necessarily (20 is not divisible by 7)
D) 8 - Not necessarily (20 is not divisible by 8)
E) 10 - Yes! Every multiple of 20 is also a multiple of 10
Process Skill: APPLY CONSTRAINTS
Since \(\mathrm{x} = 20\mathrm{m}\) for some integer m, x must be a multiple of 20, which means x must be a multiple of 10.
Let's verify with an example: If \(\mathrm{k} = 1\), then \(\mathrm{y} = 5\) and \(\mathrm{x} = \frac{20(10-1)}{3} = 60\). Check: \(3(60) + 4(5) = 180 + 20 = 200\) ✓
If \(\mathrm{k} = 4\), then \(\mathrm{y} = 20\) and \(\mathrm{x} = \frac{20(10-4)}{3} = 40\). Check: \(3(40) + 4(20) = 120 + 80 = 200\) ✓
Both 60 and 40 are multiples of 10.
The answer is E) 10.
Students might think y can only be 5, rather than understanding that y can be any multiple of 5 (like 5, 10, 15, 20, etc.). This leads them to solve only for the specific case where y = 5, missing the general pattern that x must follow.
Students may try to test random values or use trial-and-error instead of systematically expressing y as 5k and substituting into the equation. This approach becomes inefficient and may not reveal the underlying divisibility pattern for x.
Since the question asks "x must be a multiple of which of the following," students might get distracted trying to find all possible values of x instead of focusing on what x is always divisible by regardless of the specific values.
When rearranging \(3\mathrm{x} + 4\mathrm{y} = 200\) to get \(\mathrm{x} = \frac{200 - 4\mathrm{y}}{3}\), students might make sign errors or incorrect fraction operations, leading to wrong expressions for x in terms of y.
When substituting \(\mathrm{y} = 5\mathrm{k}\) into \(\mathrm{x} = \frac{200 - 4\mathrm{y}}{3}\), students might incorrectly compute \(4(5\mathrm{k})\) or make errors in simplifying \(\frac{200 - 20\mathrm{k}}{3}\), missing the factorization that leads to \(\mathrm{x} = \frac{20(10-\mathrm{k})}{3}\).
Students might not recognize that for x to be an integer, (10-k) must be divisible by 3, or they might incorrectly conclude what this means for the overall divisibility of x.
Students might test whether x could sometimes be divisible by the given options rather than checking whether x is always divisible by them. For example, they might see that some values of x are divisible by 4 and incorrectly conclude that 4 is the answer.
After determining that \(\mathrm{x} = 20\mathrm{m}\), students might not systematically check which answer choices are factors of 20. They might miss that since \(20 = 2^2 \times 5\), x must be divisible by 10, but not necessarily by 3, 6, 7, or 8.
We can solve this systematically by finding valid values that satisfy our constraints and testing what x must be divisible by.
Step 1: Set up our constraint equation
We have: \(3\mathrm{x} + 4\mathrm{y} = 200\), where y is a multiple of 5
Step 2: Choose smart values for y
Since y must be a multiple of 5, let's test \(\mathrm{y} = 20\) (chosen because \(4 \times 20 = 80\), leaving 120 for \(3\mathrm{x}\), which gives us a clean integer for x).
Step 3: Calculate corresponding x
\(3\mathrm{x} + 4(20) = 200\)
\(3\mathrm{x} + 80 = 200\)
\(3\mathrm{x} = 120\)
\(\mathrm{x} = 40\)
Step 4: Test another smart value
Let's try \(\mathrm{y} = 35\):
\(3\mathrm{x} + 4(35) = 200\)
\(3\mathrm{x} + 140 = 200\)
\(3\mathrm{x} = 60\)
\(\mathrm{x} = 20\)
Step 5: Test one more value
Let's try \(\mathrm{y} = 5\):
\(3\mathrm{x} + 4(5) = 200\)
\(3\mathrm{x} + 20 = 200\)
\(3\mathrm{x} = 180\)
\(\mathrm{x} = 60\)
Step 6: Analyze the pattern
We found that x can be 40, 20, or 60. Let's check what these numbers are all divisible by:
• \(40 = 4 \times 10\)
• \(20 = 2 \times 10\)
• \(60 = 6 \times 10\)
Step 7: Identify the common factor
All our x values (40, 20, 60) are multiples of 10. Checking against our answer choices, we see that 10 is option E.
Step 8: Verify this makes sense
Since \(4\mathrm{y}\) is always a multiple of 20 (because y is a multiple of 5), and 200 is also a multiple of 20, then \(3\mathrm{x}\) must be a multiple of 20. This means x must be a multiple of \(\frac{20}{3}\). Since x must be an integer, and \(\gcd(3,20) = 1\), x must be a multiple of 20. However, our constraint is actually that x must be a multiple of 10, which we verified with our concrete examples.