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

2007 iTest Problems/Problem 41

Revision as of 17:17, 2 July 2018 by Rockmanex3 (talk | contribs) (Better phrasing)

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 $f(n) = 2004$), $f(10^{2007}) = \boxed{2004}$.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 40 		Followed by:
Problem 42
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4

Solution

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_41&oldid=95751"