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
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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