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Solutions for Session 3, Part D
See solutions for Problems: D1 | D2 | D3 | D4 | D5 | D6| D7
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Problem D1 | |
a. | If you pick 3, the final number will be 24. |
b. | Picking 5 gives a final number of 36. |
c. | Picking 12 gives 78 |
d. | Picking 38 gives 234. |
e. | Picking n gives 6n + 6 [n >> n + 3 >> 2n + 6 >> 2n + 2 >> 6n + 6]. |
f. | Mary's original number is 7. Solving 6n + 6 = 48 gives this, or you could run 48 through the algorithm in reverse. |
g. | Joe started with 4. |
<< back to Problem D1
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Problem D2 | |
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This algorithm would undo the steps of the original by using inverse operations taken in reverse order. That would be:
• | Pick a number |
• | Divide it by 3 |
• | Add 4 |
• | Cut it in half |
• | Subtract 3 |
You should find that this algorithm undoes the one in Problem D1. Inverse algorithms like this are frequently incorporated in mathematical magic tricks.
<< back to Problem D2
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Problem D3 | |
a. | The final number will be 33. |
b. | The final number will be 51 |
c. | The final number will be 114 |
d. | The final number will be 348. |
e. | The final number will be 9n + 6 [n >> n + 3 >> 2n + 6 >> 3n + 6 >> 3n + 2 >> 9n + 6]. |
f. | This is difficult without completing D3(e), but solving 9n + 6 = 48 suggests that Mary's original number was 4 2/3. |
g. | Solving 9n + 6 = 30 shows that Joe's original number was 2 2/3. |
<< back to Problem D3
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Problem D4 | |
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It's pretty tough to draw a picture of the machine for "add the original number"! Specifically, the original number is a variable, so the results of that machine vary depending on what you put in to start. A machine labeled "+ n" might work, but notice how different that is from one that reads "+ 3".
<< back to Problem D4
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Problem D5 | |
a. | The final number will be 7. |
b. | The final number will be 7. |
c. | The final number will be 7. |
d. | The final number will be 7. |
e. | The final number will always be 7. The algorithm, starting with n, proceeds as n >> 3n >> 3n + 5 >> 6n + 10 >> 10 >> 14 >> 7. Note the step "subtract 6 times the original number" removes n completely, so the answer does not depend on n. |
f. | Since the answer is always 7, there is no way to determine what number Nancy was thinking of. |
<< back to Problem D5
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Problem D6 | |
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You almost need to know a formula for the entire algorithm before solving D3(f) and D3(g). In Problem D1, a "reverse" strategy will work, taking the algorithm step-by-step in reverse. But in Problem D3, the "add the original number" step is impossible unless you know the original number! So you could guess until you get it, or come up with a formula for the entire algorithm.
<< back to Problem D6
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Problem D7 | |
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Here's one: Pick a number, multiply it by 9, add 18, divide by 3, add 6, divide by 3 again, then subtract the original number. The answer will always be 4.
<< back to Problem D7
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