I've been puzzling over this problem for a bit. I found it on learner.org. It says, "It's a hot summer day, and Eric the Sheep is at the end of a line of sheep waiting to be shorn. There are 50 sheep in front of him. Being an impatient sort of sheep, though, every time the shearer takes a sheep from the front of the line to be shorn, Eric sneaks up two places in line. Without working out the entire problem, predict how many sheep will get shorn before Eric."
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$\begingroup$For every sheep shorn, Eric's place in line jumps forward exactly two places. So, when one sheep is removed from the line, Eric's place in line moves up one, then two, so three.
Let:
n = the number of sheep shorn
a_n = the number of sheep in front of Eric. (This is just sequence notation; you can think of it as a function a of n.) For one sheep,
a_1 = (51 - 3)
For two,
a_2 = (51 - 3) - 3 = a_1 - 3
For three,
a_3 = ((51 - 3) - 3) - 3 = a_2 - 3
For four,
a_4 = (((51 - 3) - 3) - 3) - 3 = a_2 - 3
So for any n,
a_n = a_(n-1) - 3 = 51 - 3n
Our goal in the above problem is to find the value of n that makes a_n = 0.
Therefore,
0 = 51 - 3n
51 = 3n
51/3 = 3n/3
17 = n
$\endgroup$ $\begingroup$Without working out the entire problem, every time one sheep is shorn, three sheep are removed from in front of Eric. So, we think the answer is 50 divided by 3. And we know to divide, because it's like walking 50 feet by taking 3 foot steps.
Now, 50 is not divisible by 3, but 51 is, so the answer should be about 17, without working out any details.
Looking at it closer, we may want to count Eric as the 51st sheep, which he is, but that is not exactly the right way to get exactly 17. We imagine what happens when 16 sheep have been shorn. Then 48 of the 50 sheep have been removed from in front of Eric and 2 are still standing in front of him. So, when the 17th sheep is shorn, Eric only sneaks in front of one sheep.
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