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Let \(\mathrm{s}\) be the set of all \(6\)-digit numbers of the form \(54\mathrm{x},\mathrm{y}12\), where each of \(\mathrm{x}\) and \(\mathrm{y}\) can be any of the digits \(3, 5, 8\), and \(\mathrm{x} = \mathrm{y}\) is allowed. What is the probability that a randomly chosen number from set \(\mathrm{S}\) is divisible by \(8\)?
Let's break down what we're looking for in simple terms. We have 6-digit numbers that look like 54x,y12, where the commas just help us see the structure clearly - these are actually written as 54xy12.
The key constraints are:
So our numbers look like: 543312, 543512, 543812, 545312, 545512, etc.
We need to find what fraction of these numbers are divisible by 8. Since x can take 3 values and y can take 3 values, we have \(3 \times 3 = 9\) total possible numbers in our set.
Process Skill: TRANSLATE - Converting the notation 54x,y12 into concrete examples helps us understand the structure
Here's a key insight that makes this problem much simpler: a number is divisible by 8 if and only if its last three digits form a number that's divisible by 8.
Why does this work? Think about it this way - any number can be written as:
(everything except the last 3 digits) × 1000 + (last 3 digits)
Since \(1000 = 8 \times 125\), the first part is always divisible by 8. So the entire number is divisible by 8 exactly when the last three digits are divisible by 8.
For our numbers 54xy12, the last three digits are y12. So we only need to check when y12 is divisible by 8, and we can completely ignore the 54x part!
Now let's test each possible value of y to see when y12 is divisible by 8:
When y = 3: We get 312
\(312 \div 8 = 39\) with no remainder
So 312 is divisible by 8 ✓
When y = 5: We get 512
\(512 \div 8 = 64\) with no remainder
So 512 is divisible by 8 ✓
When y = 8: We get 812
\(812 \div 8 = 101.5\)
Since this gives a decimal, 812 is NOT divisible by 8 ✗
This means that our 6-digit numbers are divisible by 8 exactly when y = 3 or y = 5.
Now we can count our favorable outcomes:
Favorable cases: Numbers where y = 3 or y = 5
Total possible cases: 9 numbers (as calculated earlier)
Probability = Favorable outcomes / Total outcomes = \(\frac{6}{9} = \frac{2}{3}\)
Process Skill: CONSIDER ALL CASES - We systematically checked each possible value of y
The probability that a randomly chosen number from set S is divisible by 8 is \(\frac{2}{3}\).
This matches answer choice (E) \(\frac{2}{3}\).
Students may incorrectly think this represents a decimal number (like 54x.y12) rather than a 6-digit integer 54xy12. This confusion could lead them to apply incorrect mathematical operations or rules.
Students might incorrectly assume that x and y must be different from each other, missing that "x = y is allowed." This would lead them to calculate only \(3 \times 2 = 6\) total numbers instead of \(3 \times 3 = 9\) total numbers.
Students may try to check divisibility by 8 using the entire 6-digit number instead of recognizing that only the last three digits (y12) matter. This makes the problem unnecessarily complex and prone to calculation errors.
When checking if 312, 512, and 812 are divisible by 8, students may make calculation mistakes. For example, incorrectly computing \(812 \div 8 = 101\) (forgetting the remainder) or making errors in the division process.
Students may correctly identify that y = 3 and y = 5 work, but then forget that for each valid y value, x can take any of the 3 possible values (3, 5, 8). They might count only 2 favorable outcomes instead of 6.
Even if students identify the correct favorable outcomes, they might use the wrong numbers in their fraction. For example, using \(\frac{2}{9}\) (thinking only 2 cases work) or \(\frac{6}{12}\) (using wrong total) instead of \(\frac{6}{9}\).
Students may correctly calculate \(\frac{6}{9}\) but fail to simplify it to \(\frac{2}{3}\), leading them to look for \(\frac{6}{9}\) among the answer choices and potentially select an incorrect option when they don't find it.