Difference between revisions of "2016 AIME I Problems/Problem 5"
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==Solution 4== | ==Solution 4== | ||
− | Notice <math>374=34\cdot 11</math> and <math>319=29\cdot 11</math>. Also, note the sum of an arithmetic series is <math>\frac{2n+k}{2} \cdot b</math>, where <math>n</math> is our first term, <math>n+k</math> is our final term, and <math>b</math> is the number of terms. Since we know both sequences of <math>n</math> and <math>t</math> have the same length, and since <math>11</math> is prime and shared by both <math>319</math> and <math>374</math>, we deduce that <math>b=11</math>. Thus from here we know <math>2n+k=68</math> and <math>2t+k=58</math> by using our other factors <math>34</math> and <math>29</math>. Finally, we add the two systems up and we get <math>2t+2k+2n=126</math>. But, notice that <math>k=b-1</math>, since the first term has <math>k=0</math>, and our last term has <math>k=b-1</math>. Plugging this back into our equation we get <math>2n+2t= | + | Notice <math>374=34\cdot 11</math> and <math>319=29\cdot 11</math>. Also, note the sum of an arithmetic series is <math>\frac{2n+k}{2} \cdot b</math>, where <math>n</math> is our first term, <math>n+k</math> is our final term, and <math>b</math> is the number of terms. Since we know both sequences of <math>n</math> and <math>t</math> have the same length, and since <math>11</math> is prime and shared by both <math>319</math> and <math>374</math>, we deduce that <math>b=11</math>. Thus from here we know <math>2n+k=68</math> and <math>2t+k=58</math> by using our other factors <math>34</math> and <math>29</math>. Finally, we add the two systems up and we get <math>2t+2k+2n=126</math>. But, notice that <math>k=b-1</math>, since the first term has <math>k=0</math>, and our last term has <math>k=b-1</math>. Plugging this back into our equation we get <math>2n+2t=106 \implies n+t=\boxed{053}</math> |
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== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=4|num-a=6}} | {{AIME box|year=2016|n=I|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:34, 25 July 2021
Problem 5
Anh read a book. On the first day she read pages in
minutes, where
and
are positive integers. On the second day Anh read
pages in
minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she completely read the
page book. It took her a total of
minutes to read the book. Find
.
Solution 1
Let be the number of days Anh reads for. Because the difference between the number of pages and minutes Anh reads for each day stays constant and is an integer,
must be a factor of the total difference, which is
. Also note that the number of pages Anh reads is
. Similarly, the number of minutes she reads for is
. When
is odd (which it must be), both of these numbers are multiples of
. Therefore,
must be a factor of
,
, and
. The only such numbers are
and
. We know that Anh reads for at least
days. Therefore,
.
Using this, we find that she reads "additional" pages and
"additional" minutes. Therefore,
, while
. The answer is therefore
.
Solution 2
We could see that both and
are divisible by
in the outset, and that
and
, the quotients, are relatively prime. Both are the
number of minutes across the
days, so we need to subtract
from each to get
and
.
Solution 3
If we let be equal to the number of days it took to read the book, the sum of
through
is equal to
Similarly,
We know that both factors must be integers and we see that the only common multiple of
and
not equal to
that will get us positive integer solutions for
and
is
. We set
so
. We then solve for
and
in their respective equations, getting
.
We also get
.
. Our final answer is
Solution 4
Notice and
. Also, note the sum of an arithmetic series is
, where
is our first term,
is our final term, and
is the number of terms. Since we know both sequences of
and
have the same length, and since
is prime and shared by both
and
, we deduce that
. Thus from here we know
and
by using our other factors
and
. Finally, we add the two systems up and we get
. But, notice that
, since the first term has
, and our last term has
. Plugging this back into our equation we get
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.