Difference between revisions of "2018 AMC 10A Problems/Problem 19"
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<cmath>7*3=1</cmath> | <cmath>7*3=1</cmath> | ||
<cmath>1*3=3</cmath> | <cmath>1*3=3</cmath> | ||
− | We see that the unit digit of <math>3^x</math> for some integer <math>x</math> will only be <math>1</math> when <math>x</math> is a multiple of <math>4</math>. Now, let's count how many numbers in <math>B</math> are divisible by <math>4</math>. This can be done by simply listing: | + | We see that the unit digit of <math>3^x</math>, for some integer <math>x</math>, will only be <math>1</math> when <math>x</math> is a multiple of <math>4</math>. Now, let's count how many numbers in <math>B</math> are divisible by <math>4</math>. This can be done by simply listing: |
<cmath>2000,2004,2008,2012,2016.</cmath> | <cmath>2000,2004,2008,2012,2016.</cmath> | ||
There are <math>5</math> numbers in <math>B</math> divisible by <math>4</math> out of the <math>2018-1999+1=20</math> total numbers. Therefore, the probability that <math>3</math> is picked from <math>A</math> and a number divisible by <math>4</math> is picked from <math>B</math> is <math>\frac{1}{5}*\frac{5}{20}=\frac{1}{20}</math>. | There are <math>5</math> numbers in <math>B</math> divisible by <math>4</math> out of the <math>2018-1999+1=20</math> total numbers. Therefore, the probability that <math>3</math> is picked from <math>A</math> and a number divisible by <math>4</math> is picked from <math>B</math> is <math>\frac{1}{5}*\frac{5}{20}=\frac{1}{20}</math>. | ||
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<cmath>3*7=1</cmath> | <cmath>3*7=1</cmath> | ||
<cmath>1*7=7</cmath> | <cmath>1*7=7</cmath> | ||
− | We see that the unit digit of <math>7^y</math> for some integer <math>y</math> will only be <math>1</math> when <math>y</math> is a multiple of <math>4</math>. This is exactly the same conditions as our last case with <math>3</math> so the probability of this case is also <math>\frac{1}{20}</math>. | + | We see that the unit digit of <math>7^y</math>, for some integer <math>y</math>, will only be <math>1</math> when <math>y</math> is a multiple of <math>4</math>. This is exactly the same conditions as our last case with <math>3</math> so the probability of this case is also <math>\frac{1}{20}</math>. |
− | Since <math>5*5=25</math> and <math>25</math> ends in <math>5</math>, the units digit of <math>5^w</math> for some integer <math>w</math> will always be <math>5</math>. Thus, the probability in this case is <math>0</math>. | + | Since <math>5*5=25</math> and <math>25</math> ends in <math>5</math>, the units digit of <math>5^w</math>, for some integer, <math>w</math> will always be <math>5</math>. Thus, the probability in this case is <math>0</math>. |
The last case we need to consider is when the number <math>9</math> is chosen from <math>A</math>. This happens with probability <math>\frac{1}{5}</math>. We list out the repeading units digit for <math>9</math> as we have done for <math>3</math> and <math>7</math>: | The last case we need to consider is when the number <math>9</math> is chosen from <math>A</math>. This happens with probability <math>\frac{1}{5}</math>. We list out the repeading units digit for <math>9</math> as we have done for <math>3</math> and <math>7</math>: | ||
<cmath>9*9=1</cmath> | <cmath>9*9=1</cmath> | ||
<cmath>1*9=9</cmath> | <cmath>1*9=9</cmath> | ||
− | We see that the units digit of <math>9^z</math> is <math>1</math> when <math>z</math> is an even number. From the <math>20</math> numbers in <math>B</math>, we see that exactly half of them are even. The probability in this case is <math>\frac{1}{5}*\frac{1}{2}=\frac{1}{10}.</math> | + | We see that the units digit of <math>9^z</math>, for some integer <math>z</math>, is <math>1</math> only when <math>z</math> is an even number. From the <math>20</math> numbers in <math>B</math>, we see that exactly half of them are even. The probability in this case is <math>\frac{1}{5}*\frac{1}{2}=\frac{1}{10}.</math> |
Finally, we can ad all of our probabilities together to get | Finally, we can ad all of our probabilities together to get | ||
<cmath>\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.</cmath> | <cmath>\frac{1}{5}+\frac{1}{20}+\frac{1}{20}+\frac{1}{10}=\boxed{\frac{2}{5}}.</cmath> | ||
~Nivek | ~Nivek |
Revision as of 15:37, 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 only care about the unit digit, 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
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
, for some integer
, is
only 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
~Nivek