Difference between revisions of "1990 AIME Problems/Problem 11"
m |
|||
(16 intermediate revisions by 5 users not shown) | |||
Line 6: | Line 6: | ||
== Solution 2 == | == Solution 2 == | ||
− | Let the largest of the | + | Let the largest of the <math>n-3</math> consecutive positive integers be <math>k</math>. Clearly <math>k</math> cannot be less than or equal to <math>n</math>, else the product of <math>n-3</math> consecutive positive integers will be less than <math>n!</math>. |
Key observation: | Key observation: | ||
− | Now for n to be maximum the smallest number (or starting number) of the | + | Now for <math>n</math> to be maximum the smallest number (or starting number) of the <math>n-3</math> consecutive positive integers must be minimum, implying that <math>k</math> needs to be minimum. But the least <math>k > n</math> is <math>n+1</math>. |
− | So the | + | So the <math>n-3</math> consecutive positive integers are <math>5, 6, 7…, n+1</math> |
− | So we have (n+1)! | + | So we have <math>\frac{(n+1)!}{4!} = n!</math> |
− | + | <math>\Longrightarrow n+1 = 24</math> | |
− | + | <math>\Longrightarrow n = 23</math> | |
+ | == Generalization == | ||
+ | Largest positive integer <math>n</math> for which <math>n!</math> can be expressed as the product of <math>n-a</math> consecutive positive integers is <math>(a+1)! - 1</math> | ||
− | + | For ex. largest <math>n</math> such that product of <math>n-6</math> consecutive positive integers is equal to <math>n!</math> is <math>7!-1 = 5039</math> | |
− | |||
− | + | Proof: | |
+ | Reasoning the same way as above, let the largest of the <math>n-a</math> consecutive positive integers be <math>k</math>. Clearly <math>k</math> cannot be less than or equal to <math>n</math>, else the product of <math>n-a</math> consecutive positive integers will be less than <math>n!</math>. | ||
− | + | Now, observe that for <math>n</math> to be maximum the smallest number (or starting number) of the <math>n-a</math> consecutive positive integers must be minimum, implying that <math>k</math> needs to be minimum. But the least <math>k > n</math> is <math>n+1</math>. | |
− | |||
− | + | So the <math>n-a</math> consecutive positive integers are <math>a+2, a+3, … n+1</math> | |
− | So | + | So we have <math>\frac{(n+1)!}{(a+1)!} = n!</math> |
+ | <math>\Longrightarrow n+1 = (a+1)!</math> | ||
+ | <math>\Longrightarrow n = (a+1)! -1</math> | ||
− | + | <math>Kris17</math> | |
− | |||
− | |||
+ | == Video Solution!!! == | ||
+ | https://www.youtube.com/watch?v=H-ZmsYjF-xE | ||
== See also == | == See also == | ||
Line 39: | Line 42: | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 09:28, 28 June 2022
Problem
Someone observed that . Find the largest positive integer
for which
can be expressed as the product of
consecutive positive integers.
Solution 1
The product of consecutive integers can be written as
for some integer
. Thus,
, from which it becomes evident that
. Since
, we can rewrite this as
. For
, we get
so
. For greater values of
, we need to find the product of
consecutive integers that equals
.
can be approximated as
, which decreases as
increases. Thus,
is the greatest possible value to satisfy the given conditions.
Solution 2
Let the largest of the consecutive positive integers be
. Clearly
cannot be less than or equal to
, else the product of
consecutive positive integers will be less than
.
Key observation:
Now for to be maximum the smallest number (or starting number) of the
consecutive positive integers must be minimum, implying that
needs to be minimum. But the least
is
.
So the consecutive positive integers are
So we have
Generalization
Largest positive integer for which
can be expressed as the product of
consecutive positive integers is
For ex. largest such that product of
consecutive positive integers is equal to
is
Proof:
Reasoning the same way as above, let the largest of the consecutive positive integers be
. Clearly
cannot be less than or equal to
, else the product of
consecutive positive integers will be less than
.
Now, observe that for to be maximum the smallest number (or starting number) of the
consecutive positive integers must be minimum, implying that
needs to be minimum. But the least
is
.
So the consecutive positive integers are
So we have
Video Solution!!!
https://www.youtube.com/watch?v=H-ZmsYjF-xE
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.