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 == | ||
Revision as of 12:32, 2 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 
digit 
times, 
digits 
times, and 
digits 
times. So if we have 
pages, we have 
digits. Since we want to have 
digits, we do 
pages less than . So, 
pages.
~Ultraman
See also
| 2018 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| 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 | ||
