Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 1"

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== Solution ==
 
== Solution ==
 
We know that we use <math>1</math> digit <math>9</math> times, <math>2</math> digits <math>90</math> times, and <math>3</math> digits <math>900</math> times. So if we have <math>999</math> pages, we have <math>1 \cdot 9 + 2 \cdot 90 + 3 \cdot 900 = 2889</math> digits. Since we want to have <math>1890</math> digits, we do <math>\frac{2889 - 1890}{3}=333</math> pages less than <math>999</math>. So, <math>999-333=\boxed {666}</math> pages.
 
We know that we use <math>1</math> digit <math>9</math> times, <math>2</math> digits <math>90</math> times, and <math>3</math> digits <math>900</math> times. So if we have <math>999</math> pages, we have <math>1 \cdot 9 + 2 \cdot 90 + 3 \cdot 900 = 2889</math> digits. Since we want to have <math>1890</math> digits, we do <math>\frac{2889 - 1890}{3}=333</math> pages less than <math>999</math>. So, <math>999-333=\boxed {666}</math> pages.
~Ultraman
 
  
 
== See also ==
 
== See also ==

Latest revision as of 17:59, 4 January 2020

Problem

A printer used 1890 digits to number all the pages in the Seripian Puzzle Book. How many pages are in the book? (For example, to number the pages in a book with twelve pages, the printer would use fifteen digits.)

Solution

We know that we use $1$ digit $9$ times, $2$ digits $90$ times, and $3$ digits $900$ times. So if we have $999$ pages, we have $1 \cdot 9 + 2 \cdot 90 + 3 \cdot 900 = 2889$ digits. Since we want to have $1890$ digits, we do $\frac{2889 - 1890}{3}=333$ pages less than $999$. So, $999-333=\boxed {666}$ pages.

See also

2018 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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