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 sheets 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
Suppose the smallest page number borrowed is and
pages are borrowed. It follows that the largest page number borrowed is
We have the following preconditions:
pages are borrowed means that
sheets are borrowed, from which
must be even.
must be odd, as the smallest page number borrowed is on the right side (odd-numbered).
- The sum of the page numbers borrowed 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 all remaining sheets will be undefined, as there will be no sheets remaining after we take
pages (
sheets) out starting from page
Therefore, the only possibility is We conclude that
pages, or
sheets, are borrowed.
~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
Solution 4
Let be pages be borrowed, the sum of the page numbers on those pages is
while the sum of the rest pages is
and we know the average of the rest is
which equals to
; multiply this out we got
and we got
. As
, we can see
and that is desired
.
~bluesoul
Solution 5
Let be the number of consecutive sheets Hiram’s roommate borrows, and let
be the number of sheets preceding the
borrowed sheets (i.e. if the friend borrows sheets
,
, and
, then
and
).
The sum of the page numbers up till sheets is
.
The last page number of the borrowed sheets would be
. Therefore, the sum of the remaining page numbers of the sheets after the
borrowed sheets would be
.
The total number of page numbers after the borrow would be .
Thus the average of the page numbers after the borrow would be:
By the arithmetic series formula, this turns out to be:
because in the changed sum, there are
numbers minus
borrowed numbers and
numbers from the first
sheets.
This simplifies to
Noticing that some terms will cancel, we expand, leading to:
Factoring, we get
The prime factorization of 325 is
. Recall that
, so
could be
,
,
, or
.
We can rule out since Hiram would have no paper left over, so the average of the page numbers he has would be
. We can now plug in the other answers for
and we if we get a valid answer for
.
. Since Hiram only has
sheets, this is clearly wrong and we can rule out
.
. Since
is
but we have only
sheets, this is also implausible so we can rule out
.
Finally, just to make sure, we test .
is an integer and
, so everything checks out. The number of consecutive sheets borrowed by Hiram’s friend is
.
~KingRavi
Solution 6
The sum of all the page numbers is
.
If we add the page numbers on each sheet, we get this sequence:
.
So we can write the sum of the numbers on the first sheet that the roommate borrowed as
for some nonnegative integer,
. If the roommate borrowed
sheets, he borrowed sheets
.
The sum of the numbers in this sequence is
.
Since there are
pages per sheet, there are
pages remaining, the average page number of the remaining sheets is
.
Therefore,
$$ (Error compiling LaTeX. Unknown error_msg)\frac{1275 - (4nk+2k^2+k)}{50-2k} = 19$%,
which simplifies to
%$ (Error compiling LaTeX. Unknown error_msg)2k^2-37k-325 = -4nk$%.
Factoring the left-hand side,
%$ (Error compiling LaTeX. Unknown error_msg)(2k+13)(k-25)=-4nk
k
k$.
In order for$ (Error compiling LaTeX. Unknown error_msg)(2k+13)(k-25)k
2k+13
k-25
k
%\frac{2k+13}{k}=2+\frac{13}{k}$$ (Error compiling LaTeX. Unknown error_msg), so
is divisible by
only if
is a factor of
.
, so
is divisible by
only if
is a factor of
.
None of the answer choices are factors of , but answer choice B is a factor of
. Hence, the answer is
.
~ azc1027
Video Solution by OmegaLearn (Arithmetic Sequences and System of Equations)
~ pi_is_3.14
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=28te8OUiVxE
~MRENTHUSIASM
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 | ||
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