Difference between revisions of "1997 PMWC Problems/Problem T3"
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| − | + | ==Solution 1== | |
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Let's call the three digit, two digit, and one digit numbers, when combined, the 0 thousands. | Let's call the three digit, two digit, and one digit numbers, when combined, the 0 thousands. | ||
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<math>20+280=300 \Longrightarrow 300*2-5=595</math> is the answer. | <math>20+280=300 \Longrightarrow 300*2-5=595</math> is the answer. | ||
| − | + | == Solution 2 == | |
Consider the numbers from 1 to 1000. Since <tt>9</tt> appears <math>\frac{1}{10}</math> of the time for each digit (if we include 0), there are <math>\frac{1000}{10} = 100</math> '<tt>9</tt>'s in each place, for a total of <math>300</math> '<tt>9</tt>'s. Repeating again up to 2000, there are <math>600</math> '<tt>9</tt>'s; we exclude <math>1998, 1999</math>, so we have <math>600 - 5 = 595</math> '<tt>9</tt>'s. | Consider the numbers from 1 to 1000. Since <tt>9</tt> appears <math>\frac{1}{10}</math> of the time for each digit (if we include 0), there are <math>\frac{1000}{10} = 100</math> '<tt>9</tt>'s in each place, for a total of <math>300</math> '<tt>9</tt>'s. Repeating again up to 2000, there are <math>600</math> '<tt>9</tt>'s; we exclude <math>1998, 1999</math>, so we have <math>600 - 5 = 595</math> '<tt>9</tt>'s. | ||
Revision as of 13:38, 20 April 2014
Problem
To type all the integers from 1 to 1997 using a typewriter on a piece of paper, how many times is the key '9' needed to be pressed?
Contents
Solution 1
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.
is the answer.
Solution 2
Consider the numbers from 1 to 1000. Since 9 appears 
of the time for each digit (if we include 0), there are 
'9's in each place, for a total of 
'9's. Repeating again up to 2000, there are 
'9's; we exclude 
, so we have 
'9's.
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 | ||
