Mathy Monday: Apples and Guards

As someone who loves math, I want to start sharing some challenging problems on my site. Most of the ones that I will share will include answers…but you’ll have to highlight the text to see the answer! I want to give you a chance to reason through it on your own. Remember: As with ANY math problem, it is possible that we will approach it in different ways. It does not mean that either of us is wrong!

appleorchard1s

A man entered an orchard that had 4 guards and picked some apples. When he left, he had to give the first guard half of his apples and one apple more. To the second guard, he had to give half of his remaining apples and one apple more. He had to do this for each of the remaining two guards. When he left the orchard, he had one apple left.

  1. How many apples did he pick at the orchard?
  2. Suppose he had encountered 5 guards instead of 4. How many apples would have been originally gathered?
  3. Suppose he had met n guards. How many apples would have been originally gathered?

Part 1 (Highlight to view)


Let An be the number of apples picked given there are n guards.

The man ended with 1 apple. 1 is one less than half of what he had approaching the fourth guard.

A0 = 1 = ½ A1 – 1

A1 = 4

Thus, approaching the fourth guard, he had 4 apples (A1) and gave 3 to the last guard. 4 is one less than half of what he had approaching the third guard.

A1 = 4 = ½ A2 – 1

A2 = 10

Thus, approaching the third guard, he had 10 apples (A2) and gave 6 to the third guard. 6 is one less than half of what he had approaching the second guard.

A2 = 6 = ½ A3 – 1

A3 = 22

Thus, approaching the second guard, he had 22 apples (A3) and gave 12 to the second guard. 12 is one less than half of what he had approaching the first guard.

A3 = 12 = ½ A4 – 1

A4 = 46

Thus, approaching the first guard, he had 46 apples (A4) and gave 24 to the first guard. So if  there were 4 guards, the man picked 46 apples.

Part 2 (Highlight to view)

Let An be the number of apples picked given there are n guards.

The man ended with 1 apple. 1 is one less than half of what he had approaching the fifth guard.

A0 = 1 = ½ A1 – 1

A1 = 4

Thus, approaching the fifth guard, he had 4 apples (A1) and gave 3 to the last guard. 4 is one less than half of what he had approaching the fourth guard.

A1 = 4 = ½ A2 – 1

A2 = 10

Thus, approaching the fourth guard, he had 10 apples (A2) and gave 6 to the fourth guard. 6 is one less than half of what he had approaching the third guard.

A2 = 6 = ½ A3 – 1

A3 = 22

Thus, approaching the third guard, he had 22 apples (A3) and gave 12 to the third guard. 12 is one less than half of what he had approaching the second guard.

A3 = 12 = ½ A4 – 1

A4 = 46

Thus, approaching the second guard, he had 46 apples (A4) and gave 24 to the second guard.  24 is one less than half of what he had approaching the first guard.

A4 = 12 = ½ A5 – 1

A5 = 94

Thus, approaching the first guard, he had 94 apples (A4) and gave 47 to the second guard. So if  there were 5 guards, the man picked 94 apples.

Part 3 (Highlight to view)

Let An be the number of apples picked given there are n guards.

A0 = 1

A1 = 1 + 3 + 6

A2 = 1 + 3 + 6 + 12

A3 = 1 + 3 + 6 + 12 + 24

A4 = 1 + 3 + 6 + 12 + 24 + 48

An = 1 + 3 + 6 + 12 + 24 + 48…

An = 1 + 3(1 + 2 + 4 + 8 + 16 +…)

 

Which is a geometric series, and can be rewritten as:

An = 1 + 3(2n – 1)

How did you do? Did you solve it in a similar way? How was your approach different? Let me know below!

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