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Learning Math Home
Patterns, Functions, and Algebra
Session 3 Part A Part B Part C Part D Part E Homework
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Session 3 Materials:



Solutions for Session 3, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7 | C8 | C9 | C10 | C11

Problem C1

The number 48 will come out at the bottom.

<< back to Problem C1


Problem C2

Try to undo the steps. The original number was 7.

<< back to Problem C2


Problem C3

The order of operations is: multiply by 2, subtract 6, divide by 10, subtract3, divide by 2. Algorithm D is the inverse of Algorithm C, so using 88 as the input for Algorithm D would answer Problem C2.

<< back to Problem C3


Problem C4

The huge Algorithm CD doesn't do anything; its output numbers will equal its input numbers.

<< back to Problem C4


Problem C5

All of them are possible in multiple ways.


59 = 10 x 5 + 9


216 = 6 x 6 x 6


15625 = 5 x 5 x 5 x 5 x 5 x 5


7280 = 9 x 9 x 9 x 10 - 10


0.12345 = [(1 / 9) / 9] x 10

<< back to Problem C5


Problem C6

The output will be 4.

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Problem C7

Use Algorithm B, which undoes Algorithm A. The input was 8.

<< back to Problem C7


Problem C8

Using Algorithm B, the input was 31.

<< back to Problem C8


Problem C9

<< back to Problem C9


Problem C10

It will leave the number unchanged, since A and B undo each other.

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Problem C11

If you can list the algorithm as a series of steps involving unchanging numerical operations (like "add 6"), then they can be undone by an algorithm which performs the inverse operation, and where the steps are performed in reverse order. Unfortunately, some operations do not have inverses, like squaring or throwing a water balloon. Think of Mr. Lewis's rule from Session 2 -- this is a rule that cannot be undone.

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