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 "1996 AHSME Problems/Problem 24"

m (→‎Solution)
(→‎Solution 3)
Line 43: Line 43:
 
Note:  If you notice that the above sums form <math>1 + 3 + 5 + 7... + (2n-1) = n^2</math>, the fact that <math>49^2</math> appears at the end should come as no surprise.
 
Note:  If you notice that the above sums form <math>1 + 3 + 5 + 7... + (2n-1) = n^2</math>, the fact that <math>49^2</math> appears at the end should come as no surprise.
  
−
==Solution 2==
+
==Solution 3==
−
We note that the number of elements in each block are
+
The <math>k</math>th appearance of 1 is at position <math>1 + 2 + \dots + k = k(k + 1)/2</math>. Then there are <math>k</math> 1's and <math>k(k + 1)/2 - k = k(k - 1)/2</math> 2's among the first <math>k(k + 1)/2</math> numbers, so the sum of these <math>k(k + 1)/2</math> terms is <math>k + k(k - 1) = k^2</math>.
  
−
<math>2, 3, 4,</math> and so on.
+
When <math>k = 49</math>, <math>k(k + 1)/2 = 1225</math>, and when <math>k = 50</math>, <math>k(k + 1)/2 = 1275</math>.
  
−
Let <math>n</math> be the number of blocks up to <math>1234</math> terms.
+
The sum of the first 1225 terms is <math>49^2 = 2401</math>. The numbers in positions 1226 through 1234 are all 2's, so their sum is <math>(1234 - 1226 + 1) \cdot 2 = 18</math>. Therefore, the sum of the first 1234 terms is <math>2401 + 18 = \boxed{2419}</math>.
−
 
 
−
We have <math>\frac {2+n+1}{2} \cdot (n)=1234</math>  
 
−
 
 
−
Solving for <math>n</math>, we get about <math>48.2</math>.
 
−
 
 
−
We have <math>48</math> full blocks and <math>1</math> partial block. We find that there are a total of <math>49</math> <math>1</math>'s
 
−
 
 
−
Now, we change every number in the sequence to <math>2</math>, and then add. We get <math>2468</math>. Since we added <math>49</math> by changing all <math>49</math> <math>1</math>'s to <math>2</math>, we must subract that <math>49</math>. Giving us
 
−
 
 
−
<math>2468-49=2419</math> <math>\boxed{B}</math>
 
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=23|num-a=25}}
 
{{AHSME box|year=1996|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:30, 17 March 2020

Problem

The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$’s separated by blocks of $2$’s with $n$ $2$’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is

$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$

Solution

The sum of the first $1$ numbers is $1$

The sum of the next $2$ numbers is $2 + 1$

The sum of the next $3$ numbers is $2 + 2 + 1$

In general, we can write "the sum of the next $n$ numbers is $1 + 2(n-1)$", where the word "next" follows the pattern established above.

Thus, we first want to find what triangular numbers $1234$ is between. By plugging in various values of $n$ into $f(n) = \frac{n(n+1)}{2}$, we find:

$f(50) = 1275$

$f(49) = 1225$

Thus, we want to add up all those sums from "next $1$ number" to the "next $49$ numbers", which will give us all the numbers up to and including the $1225^{th}$ number. Then, we can manually tack on the remaining $2$s to hit $1234$.

We want to find:

$\sum_{n=1}^{49} 1 + 2(n-1)$

$\sum_{n=1}^{49} 2n - 1$

$\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1$

$2 \sum_{n=1}^{49} n - 49$

$2\cdot \frac{49\cdot 50}{2} - 49$

$49^2$

$2401$

Thus, the sum of the first $1225$ terms is $2401$. We have to add $9$ more $2$s to get to the $1234^{th}$ term, which gives us $2419$, or option $\boxed{B}$.

Note: If you notice that the above sums form $1 + 3 + 5 + 7... + (2n-1) = n^2$, the fact that $49^2$ appears at the end should come as no surprise.

Solution 3

The $k$th appearance of 1 is at position $1 + 2 + \dots + k = k(k + 1)/2$. Then there are $k$ 1's and $k(k + 1)/2 - k = k(k - 1)/2$ 2's among the first $k(k + 1)/2$ numbers, so the sum of these $k(k + 1)/2$ terms is $k + k(k - 1) = k^2$.

When $k = 49$, $k(k + 1)/2 = 1225$, and when $k = 50$, $k(k + 1)/2 = 1275$.

The sum of the first 1225 terms is $49^2 = 2401$. The numbers in positions 1226 through 1234 are all 2's, so their sum is $(1234 - 1226 + 1) \cdot 2 = 18$. Therefore, the sum of the first 1234 terms is $2401 + 18 = \boxed{2419}$.

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1996_AHSME_Problems/Problem_24&oldid=119652"