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 |