Two Be or Not Two Be...
Out of the numbers 0 to 9,999,999 exactly how many numbers contain at least one appearance of the digit 2 ?
The answer is 5217031.
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2 solutions
@satyen nabar Note that 0 cannot be the first digit of a number, (except in the case of the number 0). Also, I believe that we need to look at the cases of 1-digit numbers, 2-digit numbers, etc., separately. With that in mind, I offered my solution to your problem. It is a bit of a surprise that numbers with at least one appearance of the digit 2 are more common! :)
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I don't think that it would matter because 10 to the power 7 refers to all the possibilities where the digits could have a value of 0 to 10 and 9 to the power of 7 also refers to all the possibilities where the 7 digits of the number could have a value of 0 to 10 leaving 2. Therefore, subtracting 10 to the power 7 from 9 to the power 7 would leave us with all the possibilities of numbers that contain at least one appearance of the digit 2.
dammit, I calculated how many numbers from 0 to 9,999,999 contain at least one digit 2 and have digits that don't repeat... :/ the answer to that question apparently is 481,203 out of 713,539 if I got it right, which is about 67.44 %. Hence the numbers with non-repeating digits and the digit 2 are more common too among the non-repeating-digit numbers.
The solution will be 10,000,000 minus the number of numbers between 0 and 9,999,999 (inclusive) that do not contain the digit 2.
To count the numbers that do not contain a 2, proceed as follows. Except for the number 0, we can't have 0 as the first digit. That leaves us with just 8 digits that can come first. There are then 9 options for each subsequent digit. So excluding the number 0 for now, we have 8 1-digit numbers, 8 * 9 2 -digit numbers, 8 * 9 2 3-digit numbers, etc., all the way up to 8 * 9 6 7-digit numbers.
So, including the number 0, we have
1 + 8 * ∑ k = 0 6 ( 9 k ) = 1 + [8 * ( 9 7 − 1 ) / ( 9 − 1 ) ] = 9 7
numbers that can be formed without the digit 2.
So the solution will be 1 0 , 0 0 0 , 0 0 0 - 9 7 = 5 2 1 7 0 3 1 .
A number that does not contain 2 is written using any of the other 9 digits: 0, 1, 3, 4, 5, 6, 7, 8, and 9.
So there are 9 choices for the first digit, then 9 choices for the second digit, and so on for each of the 7 digits. Therefore, there are 9x9x9…x9 = 9^7 = 4,782,969 numbers that do not contain the digit 2.
The remaining 10,000,000 – 4,782,969 = 5,217,031 are numbers that have at least one appearance of the digit 2.
The numbers with the digit 2 are more common!