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Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 9"

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== Solution ==
 
== Solution ==
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(a)  We know that <math>C_n</math> is a geometric series, so we can define it explicitly as follows
 +
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<math>C_n=\frac{10^n-1}{9}</math>
 +
 +
multiplying both sides by 9 yields our answer.
 +
 +
 +
(b) We have
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 +
<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>,
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<math>=(\frac{10^n+5}{3})^2</math>,
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<math>=\frac{10^{2n}}{9}+\frac{10^{n+1}}{9}+\frac{25}{9}</math>.
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 +
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>
 
(a) <math>\frac{10^n-1}{9}</math>  (b) <math>112,225</math>  (c) <math>11,122,225</math>
  

Revision as of 20:48, 17 June 2021

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) $11,122,225$

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
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