Difference between revisions of "2018 AMC 10A Problems/Problem 19"
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<math>\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5} </math> | <math>\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5} </math> | ||
− | ==Solution== | + | == Solution 1 == |
Since we only care about the unit digit, 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 <math>B</math> can be paired with <math>1</math> to make <math>1^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 only care about the unit digit, 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 <math>B</math> can be paired with <math>1</math> to make <math>1^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> | ||
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~Nivek | ~Nivek | ||
+ | |||
+ | == Solution 2 == | ||
+ | Since only the units digit is relevant, we can turn the first set into <math>\{1,3,5,7,9\}</math>. Note that <math>x^4 \equiv 1 \mod 10</math> for all odd digits <math>x</math>, except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that, <math>\mod 4</math>, this set has 5 values which correspond to <math>\{0,1,2,3\}</math>, making the probability equal for all of them. Next, check the values for which it is equal to <math>1 \mod 10</math>. There are <math>4+1+0+1+2=8</math> values for which it is equal to 1, remembering that <math>5^{4n} \equiv 1 \mod 10</math> only if <math>n=0</math>, which it is not. There are 20 values in total, and simplifying <math>\frac{8}{20}</math> gives us <math>\boxed{\frac{2}{5}}</math> or <math>\boxed{E}</math>. | ||
+ | |||
+ | <math>QED\blacksquare</math> | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2018|ab=A|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Revision as of 18:07, 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 1
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 repeating 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 add all of our probabilities together to get
~Nivek
Solution 2
Since only the units digit is relevant, we can turn the first set into . Note that
for all odd digits
, except for 5. Looking at the second set, we see that it is a set of all integers between 1999 and 2018. There are 20 members of this set, which means that,
, this set has 5 values which correspond to
, making the probability equal for all of them. Next, check the values for which it is equal to
. There are
values for which it is equal to 1, remembering that
only if
, which it is not. There are 20 values in total, and simplifying
gives us
or
.
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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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