Difference between revisions of "1997 PMWC Problems/Problem T3"
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| + | ==Problem== | ||
| + | To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times of the key '9' needed to be pressed? | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | Let's call the three digit, two digit, and one digit numbers, when combined, the 0 thousands. | ||
| + | |||
| + | The 0 thousand has the same number of nines as the one thousand, so we can compute the number of nines in the 0 thousands and multiply it by 2 and subtract 5, since we are leaving out 1998 and 1999. | ||
| + | |||
| + | one digit: one nine, obviously. | ||
| + | two digit: a nine times nine, plus 10 other nines, is 19 nines. | ||
| + | three digit: 20 per hundred, plus another hundred for the 900s, is 280. | ||
| + | |||
| + | 20+280=300. 300*2-5=595. | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{PMWC box|year=1997|num-b=T2|num-a=T4}} | ||
Revision as of 09:20, 9 October 2007
Problem
To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times of the key '9' needed to be pressed?
Solution
Let's call the three digit, two digit, and one digit numbers, when combined, the 0 thousands.
The 0 thousand has the same number of nines as the one thousand, so we can compute the number of nines in the 0 thousands and multiply it by 2 and subtract 5, since we are leaving out 1998 and 1999.
one digit: one nine, obviously. two digit: a nine times nine, plus 10 other nines, is 19 nines. three digit: 20 per hundred, plus another hundred for the 900s, is 280.
20+280=300. 300*2-5=595.
See Also
| 1997 PMWC (Problems) | ||
| Preceded by Problem T2 |
Followed by Problem T4 | |
| I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
