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:

Group of answer choices

If the value found in step 6 is 10, 20, 30, 40, 50 or 60 then the sum is even

If the last digit of the value found in step 6 is 5, then the sum is odd

If the value found in step 6 is 10, 30 or 50 then the sum is odd

If the value found in step 6 is not 20, 40, 60, 80, 100 or 120 then the sum is odd

(a) If the value found in step 6 is 10, 20, 30, 40, 50 or 60 then the sum is even
(b) If the last digit of the value found in step 6 is 5, then the sum is odd

First, we analyze the algorithm.

Assume the outcomes of the two rolls are x and y

Step 4: Calculate sum

[tex]Sum = x + y[/tex]

Step 5: Multiply Sum by 10

[tex]Result = Sum \times 10[/tex]

Step 6: Divide by 2

[tex]Result = Sum \times 10 \div 2[/tex]

[tex]Result = Sum \times 5[/tex]

This means that, the algorithm simply multiplies the sum of the rolls by 5

Option (a): If step 6 is 10, 20, 30, 40, 50 or 60, then sum is even

Make Sum the subject in [tex]Result = Sum \times 5[/tex]

[tex]Sum = \frac{Result}{5}[/tex]

When 10, 20, 30, 40, 50 and 60 are divided by 5, the result is:

[tex]Sum = 2, 4,6,8,10, 12[/tex]

All the above numbers are even number.

This means that, the sum is an even number.

Hence, (a) is true

Option (b): If the last digit of step 6 is 5, then the sum is Odd.

The possible sum of two rolls of dice, where the last digit ends in 5 are:

5, 15, 25, 35

When each of these are divided by 5, the result is:

[tex]Sum = 1,3,5,7[/tex]

It is not possible to have a sum of 1 for two rolls

So, we have:

[tex]Sum = 3,5,7[/tex]

All the above numbers are odd.

This means that, the sum is an odd number.

Hence, (b) is true

So, we can conclude that the true conditions are:

(a) and (b)