Difference between revisions of "2018 AMC 10A Problems/Problem 19"
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When a number's unit's digit is <math>9</math>, then <math>9^{2n}</math> will produce a number ending in <math>1</math>, so we have <math>20 \div 2 = 10</math> possibilities. | When a number's unit's digit is <math>9</math>, then <math>9^{2n}</math> will produce a number ending in <math>1</math>, so we have <math>20 \div 2 = 10</math> possibilities. | ||
− | Hence, we have a total of <math>5 \cdot 20 = 100</math> ways, so the probability is <math>\frac{20+5+0+5+10}{100} = \frac {40}{100} = \boxed{\textbf{(E)} ~\frac{ | + | Hence, we have a total of <math>5 \cdot 20 = 100</math> ways, so the probability is <math>\frac{20+5+0+5+10}{100} = \frac {40}{100} = \boxed{\textbf{(E)} ~\frac{2}{5}}</math>. |
~MrThinker | ~MrThinker | ||
− | ==Video Solution== | + | ==Video Solution by TheBeautyofMath== |
https://youtu.be/M22S82Am2zM?t=630 | https://youtu.be/M22S82Am2zM?t=630 | ||
~IceMatrix | ~IceMatrix | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/njyn611TJi0 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 21:13, 6 August 2024
Contents
Problem
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 units 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
~very minor edits by virjoy2001
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
.
Solution 3
By Euler's Theorem, we have that , if
.
Hence
,
work.
Also note that because
, and the latter
is clearly
. So
,
work (not counting multiples of 4 as we would be double counting if we did).
We can also note that because
, and by the same logic as why
, we are done. Hence
, and
work (not counting any of the aforementioned cases as that would be double counting).
We cannot make any more observations that add more with unit digit
, hence the number of
that have units digit one is
. And the total number of combinations of an element of the set of all
and an element of the set of all
is
. Hence the desired probability is
, which is answer choice
.
~vsamc
Solution 4 (Easy)
When a number's unit's digit is , then any power to this number will also end in
(since
for any
is always
), so we have
choices for
.
When a number's unit's digit is , then
for any
will produce a number ending with 1. So,
choices for
.
always ends in 5, so there are
possibilities for
.
When a number's unit's digit is , then this is also the same thing with
, so we have
choices.
When a number's unit's digit is , then
will produce a number ending in
, so we have
possibilities.
Hence, we have a total of ways, so the probability is
.
~MrThinker
Video Solution by TheBeautyofMath
https://youtu.be/M22S82Am2zM?t=630 ~IceMatrix
Video Solution 2
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.