Difference between revisions of "2015 AMC 12B Problems/Problem 20"
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<math>\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4</math> | <math>\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4</math> | ||
− | ==Solution | + | ==Solution== |
Simply draw a table of values of <math>f(i,j)</math> for the first few values of <math>i</math>: | Simply draw a table of values of <math>f(i,j)</math> for the first few values of <math>i</math>: | ||
Revision as of 20:54, 11 February 2018
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 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, .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.