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 "2008 UNCO Math Contest II Problems/Problem 9"

(Created page with "== Problem == Let <math>C_n = 1+10 +10^2 + \cdots + 10^{n-1}.</math> (a) Prove that <math> 9C_n = 10^n -1.</math> (b) Prove that <math>(3C_3+ 2)^2 =112225.</math> (c) Prove t...")
 
(→‎Solution 2)
 
(4 intermediate revisions by 4 users not shown)
Line 12: Line 12:
  
 
== Solution ==
 
== Solution ==
 +
(a)  We know that <math>C_n</math> is a geometric series, so we can define it explicitly as follows
  
 +
<math>C_n=\frac{10^n-1}{9}</math>
 +
 +
multiplying both sides by 9 yields our answer.
 +
 +
 +
(b) We have
 +
 +
<math>(3*111+2)^2=335^2</math>,
 +
 +
yielding <math>112,225</math>.
 +
 +
 +
(c) We say that the nth member of the sequence equals <math>(3*C_n+2)^2</math>. Expanding yields
 +
 +
<math>(3*\frac{10^n-1}{9}+2)^2</math>,
 +
 +
<math>=(\frac{10^n+5}{3})^2</math>,
 +
 +
<math>=\frac{10^{2n}}{9}+\frac{10^{n+1}}{9}+\frac{25}{9}</math>.
 +
 +
Dividing each term separately, we know that the first term will add <math>2n</math> <math>1</math>s and <math>\frac{1}{9}</math>, the second term will add <math>n+1</math> <math>1</math>s and <math>\frac{1}{9}</math>, and the third will add <math>\frac{25}{9}</math>, giving
 +
 +
<math>\overbrace{11...11}^{2n}+\overbrace{11...11}^{n+1}+\frac{27}{9}</math>,
 +
 +
<math>\overbrace{11...11}^{n-1}\overbrace{22...22}^{n}5</math>,
 +
 +
which is exactly what we wanted.
 +
 +
 +
(a) <math>\frac{10^n-1}{9}</math>  (b) <math>112,225</math>  (c) <math>11,122,225</math>
 +
 +
== Solution 2 ==
 +
 +
(a) <math>9=10-1</math>. Multiply these two binomials and we have reach our answer (remember the formula -- it's like Difference of Cubes)
 +
 +
(b)<math>C_3=111</math>. The original expression is equal to <math>\boxed{112225}</math>. (Just brute force this out).
 +
 +
(c) Now notice that each term in the sequence is <math>(3C_n)^2</math>. As seen in part (a), we see that <math>C_n=\frac{10^n-1}{9}</math>. Follow Solution 1 above.
 +
 +
~hastapasta
  
 
== See Also ==
 
== See Also ==

Latest revision as of 11:30, 27 April 2022

Problem

Let $C_n = 1+10 +10^2 + \cdots + 10^{n-1}.$

(a) Prove that $9C_n = 10^n -1.$

(b) Prove that $(3C_3+ 2)^2 =112225.$

(c) Prove that each term in the following sequence is a perfect square: \[25, 1225, 112225, 11122225, 1111222225,\ldots\]


Solution

(a) We know that $C_n$ is a geometric series, so we can define it explicitly as follows

$C_n=\frac{10^n-1}{9}$

multiplying both sides by 9 yields our answer.


(b) We have

$(3*111+2)^2=335^2$,

yielding $112,225$.


(c) We say that the nth member of the sequence equals $(3*C_n+2)^2$. Expanding yields

$(3*\frac{10^n-1}{9}+2)^2$,

$=(\frac{10^n+5}{3})^2$,

$=\frac{10^{2n}}{9}+\frac{10^{n+1}}{9}+\frac{25}{9}$.

Dividing each term separately, we know that the first term will add $2n$ $1$s and $\frac{1}{9}$, the second term will add $n+1$ $1$s and $\frac{1}{9}$, and the third will add $\frac{25}{9}$, giving

$\overbrace{11...11}^{2n}+\overbrace{11...11}^{n+1}+\frac{27}{9}$,

$\overbrace{11...11}^{n-1}\overbrace{22...22}^{n}5$,

which is exactly what we wanted.


(a) $\frac{10^n-1}{9}$ (b) $112,225$ (c) $11,122,225$

Solution 2

(a) $9=10-1$. Multiply these two binomials and we have reach our answer (remember the formula -- it's like Difference of Cubes)

(b)$C_3=111$. The original expression is equal to $\boxed{112225}$. (Just brute force this out).

(c) Now notice that each term in the sequence is $(3C_n)^2$. As seen in part (a), we see that $C_n=\frac{10^n-1}{9}$. Follow Solution 1 above.

~hastapasta

See Also

2008 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_UNCO_Math_Contest_II_Problems/Problem_9&oldid=173643"