Difference between revisions of "2018 AMC 10A Problems/Problem 19"
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− | Since we care about the unit digit only, our set <math>{11,13,15,17,19}</math> can be turned into <math>{1,3,5,7,9}</math>. Call this set <math>A</math> and call <math>{1999, 2000, 2001, \cdots , 2018}</math> set <math>B</math>. Let's do casework on the element of <math>A</math> that we choose. Since <math>1*1=1</math>, any number from from <math>B</math> can be paired with <math>1</math> to make <math>m^n</math> have a units digit of <math>1</math>. Therefore, the probability of this case happening is <math>\frac{1}{5}</math> since there is a <math>\frac{1}{5}</math> chance that the number <math>1</math> is selected from <math>A</math>. Let us consider the case where the number <math>3</math> is selected from <math>A</math>. Let's look at the unit digit when we repeatedly multiply the number <math>3</math> by itself: | + | Since we care about the unit digit only, our set <math>{ 11,13,15,17,19}</math> can be turned into <math>{1,3,5,7,9}</math>. Call this set <math>A</math> and call <math>{1999, 2000, 2001, \cdots , 2018}</math> set <math>B</math>. Let's do casework on the element of <math>A</math> that we choose. Since <math>1*1=1</math>, any number from from <math>B</math> can be paired with <math>1</math> to make <math>m^n</math> have a units digit of <math>1</math>. Therefore, the probability of this case happening is <math>\frac{1}{5}</math> since there is a <math>\frac{1}{5}</math> chance that the number <math>1</math> is selected from <math>A</math>. Let us consider the case where the number <math>3</math> is selected from <math>A</math>. Let's look at the unit digit when we repeatedly multiply the number <math>3</math> by itself: |
<cmath>3*3=9</cmath> | <cmath>3*3=9</cmath> | ||
<cmath>9*3=7</cmath> | <cmath>9*3=7</cmath> |
Revision as of 16:34, 8 February 2018
A number is randomly selected from the set
, and a number
is randomly selected from
. What is the probability that
has a units digit of
?
Solution
Since we care about the unit digit only, our set can be turned into
. Call this set
and call
set
. Let's do casework on the element of
that we choose. Since
, any number from from
can be paired with
to make
have a units digit of
. Therefore, the probability of this case happening is
since there is a
chance that the number
is selected from
. Let us consider the case where the number
is selected from
. Let's look at the unit digit when we repeatedly multiply the number
by itself:
We see that the unit digit of
for some integer
will only be
when
is a multiple of
. Now, let's count how many numbers in
are divisible by
. This can be done by simply listing:
There are
numbers in
divisible by
out of the
total numbers. Therefore, the probability that
is picked from
and a number divisible by
is picked from
is
.
Similarly, we can look at the repeating units digit for
:
We see that the unit digit of
for some integer
will only be
when
is a multiple of
. This is exactly the same conditions as our last case with
so the probability of this case is also
.
Since
and
ends in
, the units digit of
for some integer
will always be
. Thus, the probability in this case is
.
The last case we need to consider is when the number
is chosen from
. This happens with probability
. We list out the repeading units digit for
as we have done for
and
:
We see that the units digit of
is
when
is an even number. From the
numbers in
, we see that exactly half of them are even. The probability in this case is
Finally, we can ad all of our probabilities together to get