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When positive integer \(\mathrm{x}\) is divided by \(5\), the remainder is \(3\); and when \(\mathrm{x}\) is divided by \(7\), the remainder is \(4\). When positive integer \(\mathrm{y}\) is divided by \(5\), the remainder is \(3\); and when \(\mathrm{y}\) is divided by \(7\), the remainder is \(4\). If \(\mathrm{x} > \mathrm{y}\), which of the following must be a factor of \(\mathrm{x} - \mathrm{y}\)?
Let's understand what we're dealing with in everyday language. We have two positive integers, x and y, that both follow the exact same pattern when divided by 5 and 7:
• When either number is divided by 5, there's always 3 left over
• When either number is divided by 7, there's always 4 left over
Think of this like having two different bags of marbles that both follow the same packing rules. If you group the marbles from either bag into groups of 5, you'll always have 3 marbles left over. If you group them into groups of 7, you'll always have 4 left over.
We need to find what must always divide the difference between these two numbers when \(\mathrm{x} > \mathrm{y}\).
Process Skill: TRANSLATE - Converting the remainder language into a clear mathematical relationship
Now let's write these numbers in a way that shows their structure clearly.
Since x leaves remainder 3 when divided by 5, we can write:
\(\mathrm{x} = 5\mathrm{k} + 3\) for some integer k
Since x leaves remainder 4 when divided by 7, we can write:
\(\mathrm{x} = 7\mathrm{m} + 4\) for some integer m
Similarly for y, since it follows the exact same remainder pattern:
\(\mathrm{y} = 5\mathrm{p} + 3\) for some integer p
\(\mathrm{y} = 7\mathrm{q} + 4\) for some integer q
The key insight here is that both x and y must satisfy both remainder conditions simultaneously.
Here's where the magic happens. Since both x and y satisfy the exact same remainder conditions, they must both belong to the same "family" of numbers.
Let's find what this family looks like by finding the smallest positive number that satisfies both conditions:
• Remainder 3 when divided by 5: ...8, 13, 18, 23, 28, 33...
• Remainder 4 when divided by 7: ...11, 18, 25, 32...
The number 18 appears in both lists! Let's verify:
• \(18 \div 5 = 3\) remainder \(3\) ✓
• \(18 \div 7 = 2\) remainder \(4\) ✓
Now here's the crucial pattern: once we find one number that works (18), all other numbers that satisfy both conditions differ from 18 by multiples of \(5 \times 7 = 35\).
This means both x and y can be written as:
\(\mathrm{x} = 18 + 35\mathrm{r}\) for some integer r
\(\mathrm{y} = 18 + 35\mathrm{s}\) for some integer s
Process Skill: INFER - Recognizing that numbers with identical remainder conditions differ by multiples of the product of the divisors
Now we can find the difference:
\(\mathrm{x} - \mathrm{y} = (18 + 35\mathrm{r}) - (18 + 35\mathrm{s})\)
\(\mathrm{x} - \mathrm{y} = 18 + 35\mathrm{r} - 18 - 35\mathrm{s}\)
\(\mathrm{x} - \mathrm{y} = 35\mathrm{r} - 35\mathrm{s}\)
\(\mathrm{x} - \mathrm{y} = 35(\mathrm{r} - \mathrm{s})\)
This shows us that \(\mathrm{x} - \mathrm{y}\) must always be a multiple of 35, regardless of what specific values x and y have (as long as they satisfy the remainder conditions).
Therefore, 35 must be a factor of \(\mathrm{x} - \mathrm{y}\).
Looking at our answer choices:
The answer is E. 35 must be a factor of \(\mathrm{x} - \mathrm{y}\) because both x and y belong to the same arithmetic sequence with common difference 35, so their difference is always a multiple of 35.
1. Misunderstanding what "must be a factor" means
Students often confuse "must be a factor" with "could be a factor." They might try to find specific values of x and y and check which answer choices work for those particular cases, rather than understanding that we need something that works for ALL possible pairs of x and y satisfying the given conditions. This leads them to test individual examples instead of finding the general relationship.
2. Not recognizing that identical remainder conditions create a pattern
Many students get stuck trying to solve the individual remainder equations (\(\mathrm{x} \equiv 3 \pmod{5}\) and \(\mathrm{x} \equiv 4 \pmod{7}\)) without realizing that since both x and y satisfy the exact same remainder conditions, they must belong to the same arithmetic sequence. They might attempt complex algebraic manipulations instead of recognizing the simpler pattern-based approach.
3. Forgetting to use the Chinese Remainder Theorem concept
Students may not realize that when a number satisfies multiple remainder conditions simultaneously, all such numbers differ by multiples of the least common multiple of the divisors. They might try to work with each remainder condition separately instead of finding the combined pattern.
1. Arithmetic errors when finding the base number
When finding the smallest positive number that satisfies both remainder conditions (18 in this case), students often make calculation errors. They might incorrectly verify that \(18 \div 5 = 3\) remainder \(3\) or \(18 \div 7 = 2\) remainder \(4\), leading them to use the wrong base number for their calculations.
2. Incorrectly calculating the common difference
Students might forget that the common difference should be LCM(5,7) = 35, not \(5 \times 7 = 35\) by coincidence. Some might incorrectly use \(5 + 7 = 12\) as the common difference, thinking that remainders add up, which would lead to completely wrong expressions for x and y.
3. Algebraic manipulation errors in the final subtraction
When calculating \(\mathrm{x} - \mathrm{y} = (18 + 35\mathrm{r}) - (18 + 35\mathrm{s})\), students often make sign errors or forget to distribute the negative sign properly. They might get \(\mathrm{x} - \mathrm{y} = 35\mathrm{r} + 35\mathrm{s}\) instead of \(\mathrm{x} - \mathrm{y} = 35(\mathrm{r} - \mathrm{s})\), leading to incorrect conclusions about the factors.
1. Choosing a factor of 35 instead of 35 itself
After correctly determining that \(\mathrm{x} - \mathrm{y} = 35(\mathrm{r} - \mathrm{s})\), students might notice that \(35 = 5 \times 7\) and incorrectly conclude that either 5 or 7 alone must be factors. While these are indeed factors, the question asks what "must be" a factor, and since neither 5 nor 7 appears in the answer choices, they might get confused and pick an unrelated option.
2. Second-guessing the mathematical result
Some students correctly arrive at the conclusion that 35 must be a factor but then doubt their work because 35 seems "too obvious" or "too large." They might change their answer to a smaller number like 12 or 15, thinking they made an error in their reasoning when they actually solved it correctly.
Step 1: Find specific values that satisfy both remainder conditions
We need numbers that leave remainder 3 when divided by 5 and remainder 4 when divided by 7.
Numbers with remainder 3 when divided by 5: 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53...
Let's check which of these also give remainder 4 when divided by 7:
Step 2: Choose smart numbers
Let \(\mathrm{x} = 53\) and \(\mathrm{y} = 18\) (both satisfy the conditions and \(\mathrm{x} > \mathrm{y}\))
Verify: \(53 \div 5 = 10\) remainder \(3\) ✓ and \(53 \div 7 = 7\) remainder \(4\) ✓
Verify: \(18 \div 5 = 3\) remainder \(3\) ✓ and \(18 \div 7 = 2\) remainder \(4\) ✓
Step 3: Calculate the difference
\(\mathrm{x} - \mathrm{y} = 53 - 18 = 35\)
Step 4: Test which answer choice must be a factor
Step 5: Verify with another pair
Let's confirm with \(\mathrm{x} = 88\), \(\mathrm{y} = 18\): \(\mathrm{x} - \mathrm{y} = 70\), and \(70 \div 35 = 2\) ✓
The answer is E. 35