Difference between revisions of "2007 iTest Problems/Problem 41"

Revision as of 17:17, 2 July 2018

Problem

The sequence of digits $123456789101112131415161718192021\ldots$ is obtained by writing the positive integers in order. If the $10^{nth}$ digit in this sequence occurs in the part of the sequence in which the m-digit numbers are placed, define $f(n)$ to be $m$ . For example, $f(2) = 2$ because the $100^{\text{th}}$ digit enters the sequence in the placement of the two-digit integer $55$ . Find the value of $f(2007)$ .

Solution

To begin, we can find out for what values of $n$ result in a certain $f(n)$ (or $m$ ). A table can help organize the findings.

$m$	Number of digits that make up $m$ -digit numbers	Highest value of $n$ where $f(n) = m$
$1$	$9$	$9$
$2$	$180$	$189$
$3$	$2700$	$2889$
$4$	$36000$	$38889$
$5$	$45 \cdot 10^4$	$488889$

It seems as if for a given value $m$ , highest $n$ where $f(n) = m$ is a string of digits with $n-1$ followed by $n-1$ eights and a nine. To prove that this is the case, we use induction. The base case for both of our wanted outcomes is seen in the table.

For the inductive step, assume the highest $n$ where $f(n) = m$ is the string of digits $(m-1) \underbrace{888 \cdots 8}_{m-1 \text{ digits}} 9$ . The number of digits that make up $(m+1)$ digit numbers is $((10^{m+1} - 1) - (10^{m} + 1)) \cdot (m+1) = 10^{m} \cdot 9(m+1)$ , which is the string of digits $9(m+1)$ followed by $m$ zeroes.

Note that the last $m$ digits of $10^{m} \cdot 9m$ are zeroes, so the last $m$ digits are still $\underbrace{888 \cdots 8}_{m-1 \text{ digits}} 9$ . The preceding string of digits are $m-1+9m+9 = 10m + 8$ . The first string of digits are the string of digits of $m$ , and it is followed by $m$ eights and a nine, so the inductive step is complete.

Using the new discovery, the highest $n$ where $f(n) = 2007$ has $4 + 2006 + 1 = 2011$ digits, the highest $n$ where $f(n) = 2003$ has $4 + 2002 + 1 = 2007$ digits, and the highest $n$ where $f(n) = 2004$ has $4 + 2003 + 1 = 2008$ digits. Since $10^{2007}$ has $2008$ digits and $10^{2007}$ is less than $2003\underbrace{888 \cdots 8}_{2003 \text{ digits}} 9$ (highest $n$ where $f(n) = 2004$ ), $f(10^{2007}) = \boxed{2004}$ .

Solution

@@ Line 37: / Line 37: @@
 <br>
-For the inductive step, assume the highest <math>n</math> where <math>f(n) = m</math> is the string of digits <math>(m-1) \underbrace{888 \cdots 8}_{m-1 \text{ digits}} 9</math>.
+For the inductive step, assume the highest <math>n</math> where <math>f(n) = m</math> is the string of digits <math>(m-1) \underbrace{888 \cdots 8}_{m-1 \text{ digits}} 9</math>.  The number of digits that make up <math>(m+1)</math> digit numbers is <math>((10^{m+1} - 1) - (10^{m} + 1)) \cdot (m+1) = 10^{m} \cdot 9(m+1)</math>, which is the string of digits <math>9(m+1)</math> followed by <math>m</math> zeroes.
-There are <math>((10^{m+1} - 1) - (10^{m} + 1)) \cdot (m+1) = 10^{m} \cdot 9(m+1)</math> (the string of digits <math>9(m+1)</math> followed by <math>m</math> zeroes) digits that make up <math>(m+1)</math> digit numbers.
 <br>

2007 iTest (Problems, Answer Key)
Preceded by: Problem 40	Followed by: Problem 42
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Difference between revisions of "2007 iTest Problems/Problem 41"

Revision as of 17:17, 2 July 2018

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