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Solutions for Session 3, Part C
See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7 | C8 | C9 | C10 | C11
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Problem C3 | |
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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
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Problem C4 | |
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The huge Algorithm CD doesn't do anything; its output numbers will equal its input numbers.
<< back to Problem C4
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Problem C5 | |
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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
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Problem C7 | |
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Use Algorithm B, which undoes Algorithm A. The input was 8.
<< back to Problem C7
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Problem C10 | |
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It will leave the number unchanged, since A and B undo each other.
<< back to Problem C10
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Problem C11 | |
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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.
<< back to Problem C11
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