Difference between revisions of "2017 AMC 10B Problems/Problem 14"
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In <math>4</math> out of the <math>5</math> cases, the result was <math>1 \pmod 5</math>, and since each case occurs equally as <math>2020 \equiv 0 \pmod 5</math>, the answer is <math>\boxed{\textbf{(D) }\frac{4}{5}}</math> | In <math>4</math> out of the <math>5</math> cases, the result was <math>1 \pmod 5</math>, and since each case occurs equally as <math>2020 \equiv 0 \pmod 5</math>, the answer is <math>\boxed{\textbf{(D) }\frac{4}{5}}</math> | ||
+ | ==Video Solution== | ||
+ | https://youtu.be/Oj3Z1JhvoiE | ||
+ | |||
+ | ~savannahsolver | ||
{{AMC10 box|year=2017|ab=B|num-b=13|num-a=15}} | {{AMC10 box|year=2017|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:45, 11 January 2021
Problem
An integer is selected at random in the range
. What is the probability that the remainder when
is divided by
is
?
Solution 1
Notice that we can rewrite as
. By Fermat's Little Theorem, we know that
if
. Therefore for all
we have
. Since
, and
is divisible by
,
of the possible
are divisible by
. Therefore,
with probability
or
.
Solution 2
Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits .
The pattern for
is
, no matter what power, so
doesn't work. Likewise, the pattern for
is always
. Doing the same for the rest of the digits, we find that the units digits of
,
,
,
,
,
,
and
all have the remainder of
when divided by
, so
.
Solution 3 (Casework)
We can use modular arithmetic for each residue of
If , then
If , then
If , then
If , then
If , then
In out of the
cases, the result was
, and since each case occurs equally as
, the answer is
Video Solution
~savannahsolver
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 10 Problems and Solutions |
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