Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AIME II Problems/Problem 12"

(Created blank page)
 
 
(7 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 +
==Problem==
 +
For <math>n \ge 1</math> call a finite sequence <math>(a_1, a_2 \ldots a_n)</math> of positive integers <i>progressive</i> if <math>a_i < a_{i+1}</math> and <math>a_i</math> divides <math>a_{i+1}</math> for all <math>1 \le i \le n-1</math>. Find the number of progressive sequences such that the sum of the terms in the sequence is equal to <math>360</math>.
  
 +
==Solution==
 +
If the first term is <math>x</math>, then dividing through by <math>x</math>, we see that we can find the number of progressive sequences whose sum is <math>\frac{360}{x} - 1</math>, and whose first term is not 1. If <math>a(k)</math> denotes the number of progressive sequences whose sum is <math>k</math> and whose first term is not 1, then we can express the answer <math>N</math> as follows:
 +
 +
<cmath>\begin{align*}N &= a(359) + a(179) + a(119) + a(89) + a(71) + a(59) + a(44) + a(39) \\
 +
&+ a(35) + a(29) + a(23) + a(19) + a(17) + a(14) + a(11) + a(9) \\
 +
&+ a(8) + a(7) + a(5) + a(4) + a(3) + a(2) + a(1) + 1
 +
\end{align*}</cmath>
 +
 +
And in general, we have <math>a(k) = 1+\sum_{d|k\text{ and } d\ne 1,k}a(d-1)</math>.
 +
 +
The <math>+1</math> at the end accounts for the sequence whose only term is 360. Fortunately, many of these numbers are prime; we have <math>a(p) = 1</math> for primes <math>p</math> as the only such sequence is "<math>p</math>" itself. Also, <math>a(1) = 0</math>. So we have
 +
 +
<cmath> N = 15 + a(119) + a(44) + a(39) + a(35) + a(14) + a(9) + a(8) + a(4) </cmath>
 +
 +
For small <math>k</math>, <math>a(k)</math> is easy to compute: <math>a(4) = 1</math>, <math>a(8) = 2</math>, <math>a(9) = 2</math>. For intermediate <math>k</math> (e.g. <math>k=21</math> below), <math>a(k)</math> can be computed recursively using previously-computed values of <math>a(\cdot)</math>, similar to dynamic programming. Then we have
 +
<cmath>\begin{align*}
 +
a(14) &= 1+a(6) = 1+2 = 3\\
 +
a(35) &= 1+a(6)+a(4) = 1 + 2 + 1 = 4 \\
 +
a(39) &= 2 + a(12) = 2+4 = 6 \\
 +
a(44) &= 2 + a(21) + a(10) = 2+4+2=8 \\
 +
a(119) &= 1 + a(16) + a(6) = 1 + 3 + 2 = 6
 +
\end{align*}</cmath>
 +
Thus the answer is <math>N = 15+6+8+6+4+3+2+2+1 = \boxed{47}</math>.
 +
 +
-scrabbler94
 +
 +
==Video Solution==
 +
https://youtu.be/T-8-B-XgiiI
 +
 +
~MathProblemSkills.com
 +
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=9Zfi2AVq6ak ~ MathEx
 +
 +
==See Also==
 +
{{AIME box|year=2019|n=II|num-b=11|num-a=13}}
 +
{{MAA Notice}}

Latest revision as of 09:20, 20 November 2024

Problem

For $n \ge 1$ call a finite sequence $(a_1, a_2 \ldots a_n)$ of positive integers progressive if $a_i < a_{i+1}$ and $a_i$ divides $a_{i+1}$ for all $1 \le i \le n-1$. Find the number of progressive sequences such that the sum of the terms in the sequence is equal to $360$.

Solution

If the first term is $x$, then dividing through by $x$, we see that we can find the number of progressive sequences whose sum is $\frac{360}{x} - 1$, and whose first term is not 1. If $a(k)$ denotes the number of progressive sequences whose sum is $k$ and whose first term is not 1, then we can express the answer $N$ as follows:

\begin{align*}N &= a(359) + a(179) + a(119) + a(89) + a(71) + a(59) + a(44) + a(39) \\ &+ a(35) + a(29) + a(23) + a(19) + a(17) + a(14) + a(11) + a(9) \\ &+ a(8) + a(7) + a(5) + a(4) + a(3) + a(2) + a(1) + 1 \end{align*}

And in general, we have $a(k) = 1+\sum_{d|k\text{ and } d\ne 1,k}a(d-1)$.

The $+1$ at the end accounts for the sequence whose only term is 360. Fortunately, many of these numbers are prime; we have $a(p) = 1$ for primes $p$ as the only such sequence is "$p$" itself. Also, $a(1) = 0$. So we have

\[N = 15 + a(119) + a(44) + a(39) + a(35) + a(14) + a(9) + a(8) + a(4)\]

For small $k$, $a(k)$ is easy to compute: $a(4) = 1$, $a(8) = 2$, $a(9) = 2$. For intermediate $k$ (e.g. $k=21$ below), $a(k)$ can be computed recursively using previously-computed values of $a(\cdot)$, similar to dynamic programming. Then we have \begin{align*} a(14) &= 1+a(6) = 1+2 = 3\\ a(35) &= 1+a(6)+a(4) = 1 + 2 + 1 = 4 \\ a(39) &= 2 + a(12) = 2+4 = 6 \\ a(44) &= 2 + a(21) + a(10) = 2+4+2=8 \\ a(119) &= 1 + a(16) + a(6) = 1 + 3 + 2 = 6 \end{align*} Thus the answer is $N = 15+6+8+6+4+3+2+2+1 = \boxed{47}$.

-scrabbler94

Video Solution

https://youtu.be/T-8-B-XgiiI

~MathProblemSkills.com


Video Solution

https://www.youtube.com/watch?v=9Zfi2AVq6ak ~ MathEx

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_12&oldid=235142"