a. Write down a secret number, which can be any whole number
Multiply by 5
Add 2
Multiply by 4
Add 2
Multiply by 5
Subtract 50
Drop the two zeros at the end.
Do you recognize the result?
B. Let S stand for a secret number that you begin with in part (a). Write an expression in terms of S that shows the result of applying the steps in part (a). Apply properties of arithmetic to this expression to explain why the end result of the steps in part (a) is always predictable.
{[(5s+2)4]+2}5 - 50
{20s+8+2}5 - 50
100 s + 50 - 50
100 s
drop two zeros and get s
17
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mjk
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A. I recognize the result as the original secret number. This mathematical operation is called a reversal operation, where you perform a series of arithmetic steps and eventually return to the original value.
B. Let's break down the steps in part (a) and express them in terms of S, the secret number:
1. Multiply by 5: 5S
2. Add 2: 5S + 2
3. Multiply by 4: 4(5S + 2) = 20S + 8
4. Add 2: 20S + 8 + 2 = 20S + 10
5. Multiply by 5: 5(20S + 10) = 100S + 50
6. Subtract 50: 100S + 50 - 50 = 100S
Simplifying the expression, we can see that the end result of applying the steps in part (a) is always 100S, which is a multiple of the original secret number. Therefore, the end result is always predictable because it is a direct multiple of the secret number.
To explain this predictability using properties of arithmetic, let's consider the distributive property and the additive identity property:
By using the distributive property, we can split the expression into two terms:
100S + 50 = 100S + (50 - 50)
Applying the additive identity property, we have:
100S + (50 - 50) = 100S + 0
Since adding 0 to any value does not change its value, we can simplify the expression as:
100S + 0 = 100S
Therefore, we can observe that the end result is always 100S, regardless of the initial secret number. This predictability arises from the properties of arithmetic and the specific steps involved in the given operations.