2015 AMC 12B Problems/Problem 20
Contents
Problem
For every positive integer , let
be the remainder obtained when
is divided by 5. Define a function
recursively as follows:
What is ?
Solution
Simply take some time to draw a table of values of for the first few values of
:
Now we claim that for ,
for all values
. We will prove this by induction on
and
. The base cases for
, have already been proven.
For our inductive step, we must show that for all valid values of ,
if for all valid values of
,
.
We prove this itself by induction on . For the base case,
,
. For the inductive step, we need
if
. Then,
by our inductive hypothesis from our inner induction and
from our outer inductive hypothesis. Thus,
, completing the proof.
It is now clear that for ,
for all values
.
Thus, .
Solution 2
We are given that
Then,
. Thus
. Since
, we get
Then,
. Thus
. Since
, we get
Now
. Thus
Adding them all up we get
This means that
,
,
,
, and
. Thus
never takes the value 2.
Since , this implies that
for any
. By induction,
for any
. It follows that
for any
. Thus
only takes values in
. In fact, it alternates between 1 and 3:
, then
, then
, and so on.
Repeating the argument above, we see that can only take values in
. However,
for any
implies that
for any
. Thus
for all
. We can easily verify this:
, then
, then
, and so on.
Then . Moreover,
for all
. Continuing in this manner we see that
for all
.
In particular, Thus, .
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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