The algorithm below simulates rolling a regular 6-sided die twice. Consider the goal of determining if the sum of the values of the two rolls is odd or even.

Step 1: Using a random number generator, get a value between
1 - 6 inclusively
Step 2: Remember that number
Step 3: Repeat steps 1 and 2
Step 4: Add the two remembered numbers together
Step 5: Multiply that sum by 10
Step 6: Divide the result of that multiplication by 2

What conclusions can be made with regard to what we are trying to determine? Select two answers:

Question

The algorithm below simulates rolling a regular 6-sided die twice. Consider the goal of determining if the sum of the values of the two rolls is odd or even.

Step 1:        Using a random number generator, get a value between 
               1 - 6 inclusively
Step 2:        Remember that number
Step 3:        Repeat steps 1 and 2
Step 4:        Add the two remembered numbers together
Step 5:        Multiply that sum by 10
Step 6:        Divide the result of that multiplication by 2

What conclusions can be made with regard to what we are trying to determine? Select two answers:

BlackTom AI · Accepted Answer

We start by restating the question and the available answer choices to ensure clarity before digging into each option.
Question: The described algorithm simulates rolling a six-sided die twice and asks what conclusions can be drawn about whether the sum is odd or even. The two remembered rolls are summed, then multiplied by 10, and finally divided by 2. Which two statements about the parity of the sum can be concluded?
Options to evaluate:
A) If the value found in step 6 is not 20, 40, 60, 80, 100 or 120 then the sum is odd
B) If the value found in step 6 is 10, 20, 30, 40, 50 or 60 then the sum is even
C) If the last digit of the value found in step 6 is 5, then the sum is odd
D) If the value found in step 6 is 10, 30 or 50 then the sum is odd

Now, examine each option carefully and connect it to how parity behaves through the steps.
Option A:
- The sum of two dice can be even or odd. After summing, you multiply by 10 and then divide by 2, which is equivalent to multiplying the sum by 5. The result from step 6 equals 5 × (sum of the two rolls).
- If the original sum is even, 5 × (even) yields an even number in step 6; if the original sum is odd, 5 × (odd) yields an odd number in step 6.
- The even outcomes for step 6 are 20, 40, 60, 80, 100, 120 (corresponding to sum values 4, 8, 12, 16, 20, 24 if you tracked that route, but in our scenario sums only go 2–12, so the actual even step-6 values are 20, 40, 60, which map to sums 4, 8, 12; the key point is these are the even results).
- Therefore, if step 6 yields a value that is not one of these even values, the sum must be odd. This makes A a correct statement about the parity.
Option B:
- This asserts that if step 6 is any of 10, 20, 30, 40, 50, or 60, then the sum is even. However, among these values, 10, 20, 30, 40, 50, 60 include both even results (20, 40, 60) and the odd results (10, 30, 50 are not actually produced by the described mapping of sum × 5). More importantly, the actual parity mapping is: even step-6 values correspond to even sums, and odd step-6 values correspond to odd sums. The inclusion of 10, 30, or 50 as step-6 values would incorrectly imply parity for a case that depends on the original sum. Hence the claim is not reliably correct as stated. This makes B dubious or incorrect as a universal rule.
Option C:
- The step-6 value ends with digit 5 exactly when the sum is odd, because 5 × an odd sum yields an odd multiple of 5, which ends with 5. Conversely, even sums yield step-6 values ending with 0 (since they are multiples of 10). Therefore, seeing a last digit of 5 guarantees the sum is odd. This is a correct and robust statement about parity.
Option D:
- This asserts that step 6 values of 10, 30, or 50 imply the sum is odd. But as noted, these values correspond to even sums (they are multiples of 10), not odd sums. Therefore D is incorrect.

In summary, Option A correctly captures the conditional that the sum being odd corresponds to step 6 not yielding an even multiple (20, 40, 60, 80, 100, 120). Option C correctly ties the last digit 5 to an odd sum. Option B is not reliably correct because it mixes specific step-6 values in a way that does not consistently map to even sums, and Option D incorrectly asserts odd-sum parity for 10, 30, or 50. The two statements that align with the parity behavior of the described process are A and C.

(72ab92) AP CS Principles 1.2 AP-Style MC Practice

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