2021 AMC 10A Problems/Problem 22
Contents
Problem
Hiram's algebra notes are pages long and are printed on
sheets of paper; the first sheet contains pages
and
, the second sheet contains pages
and
, and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly
. How many sheets were borrowed?
Solution 1
Suppose the roommate took pages through
, or equivalently, page numbers
through
. Because there are
numbers taken,
The first possible solution that comes to mind is if
, which indeed works, giving
and
. The answer is
~Lcz
Solution 2 (Different Variable Choice, Similar Logic)
Suppose the smallest page number removed is and
pages are removed. It follows that the largest page number removed is
We have the following preconditions:
pages are removed means that
sheets are removed, from which
must be even.
must be odd, as the smallest page number removed is on the right side (odd-numbered).
- The sum of the page numbers removed is
Together, we have
The factors of
are
Since
is even, we only have a few cases to consider:
Since only
are possible:
If then there will not be sufficient pages when we take
pages out starting from page
If then the average page number of the remaining sheets will be undefined, as there will be no sheets remaining after we take all
pages (
pages) out starting from page
Therefore, the only possibility is from which
pages, or
sheets, are taken out.
~MRENTHUSIASM
Solution 3
Let be the number of sheets borrowed, with an average page number
. The remaining
sheets have an average page number of
which is less than
, the average page number of all
pages, therefore
. Since the borrowed sheets start with an odd page number and end with an even page number we have
. We notice that
and
.
The weighted increase of average page number from to
should be equal to the weighted decrease of average page number from
to
, where the weights are the page number in each group (borrowed vs. remained), therefore
Since we have either
or
. If
then
. If
then
which is impossible. Therefore the answer should be
~asops
Video Solution by OmegaLearn (Arithmetic Sequences and System of Equations)
~ pi_is_3.14
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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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