What percent of numbers (ie, counting numbers, 1, 2, 3, 4, … and keep going) contain the digit '4'?

(I will look for my source, but I think this has been around a long time)

1. May 9, 2006 am31 6:51 am 6:51 am

Is it 11%? My thought is that the set of n-digit numbers can be created by taking 9 copies of the set of (n-1) digit numbers and prepending a number (1 through 9) to each element of each set. The set of n-digit numbers beginning with 4 is as large as the entire (n-1) digit set, and is 1/9 of the n-digit set. Is this valid? Is there a way to explain it more clearly?

2. May 9, 2006 am31 11:23 am 11:23 am

11 of the first 100 hundred numbers contain the digit 4, and of the second hundred, and of the third, but 100 of the next 100 numbers contain a “4.” 11% is too low.

There is an error in “The set of n-digit numbers beginning with 4 is as large as the entire (n-1) digit set”

Try n = 2, n – 1 = 1.

3. May 10, 2006 am31 2:27 am 2:27 am

Am I missing something?
4, 14, 24, 34, 44, 54, 64, 74, 84, 94, 40, 41, 42, 43, 45, 46, 47, 48, 49
That’s 19 numbers of the first 100 that contain a 4.

4. May 10, 2006 am31 2:34 am 2:34 am

No, that was my mistake. 19 of the first 100, 19 of the next 100, 19 of the next hundred, and 100 of the next 100. I was so focused on the underestimate that I lost sight of the actual numbers. Ouch.

5. May 10, 2006 am31 2:53 am 2:53 am

0-10 = 1
0-100 = 40-49 (10) + 9 sets of 1 (9) = 19
0-1000 = 400-499 (100) + 9 sets of 19 (171) = 271
0-10000 = 4000-4999 (1000) + 9 sets of 271 (2439) = 3439
0-100000 = 40000-49999 (10000) + 9 sets of 3439 (30951) = 40951
0-1000000 = 400000-499999 (100000) + 9 sets of 40951 (368559) = 468559

out of 10^n numbers, the number that contain at least one digit 4 equals 10^(n-1) + 9*10^(n-2) + 9*10^(n-3) and so on?

6. May 10, 2006 am31 2:59 am 2:59 am

Um, never mind.

7. May 10, 2006 am31 3:09 am 3:09 am

out of 10^n numbers, the number that contain at least one digit 4 equals (9^0)(10^(n-1)) + (9^1)(10^(n-2)) + (9^2)(10^(n-3)) and so on?

even if this is right, which it probably isn’t (it’s late & I’m sick) it’s not exactly a percent, is it? *sigh*

8. May 10, 2006 am31 3:10 am 3:10 am

ms. frizzle,
yes
though for each group you might also count the number without 4’s and subtract. For example, from 1 – 1000 there are 1000 numbers, of which 9*9*9 do not include a ‘4’ (think, instead of 10 options for each digit, we only have 9). 9*9*9 = 729, and 1000 – 729 = 271. And so on.

9. May 10, 2006 am31 3:21 am 3:21 am

ms frizzle,
you are too close. I’ll let you finish, or let someone else step in.

Remember to round your answer to the nearest whole percent. : )

10. May 13, 2006 am31 1:22 am 1:22 am

Yes, ms. frizzle is *so* close!

Here is a related puzzle. Is the sum of the reciprocals of the positive integers that do not contain a ‘4’ bounded or unbounded? That is, does the infinite sum…

1/1 + 1/2 + 1/3 + 1/5 + … + 1/13 + 1/15 + … + 1/23 + 1/25 + … + 1/33 + 1/35 + … + 1/39 + 1/50 + …

… have a (finite) limit?

How about the sum of the reciprocals of the positive integers that *do* contain a ‘4’? Is that bounded or unbounded?

11. May 13, 2006 pm31 11:00 pm 11:00 pm

I originally found this puzzle at The Grey Labyrinth, a puzzles site

12. June 9, 2006 am30 1:36 am 1:36 am

100% of the numbers contain a 4 (after rounding). Frizzle was close, but it needs to be extended.

When you get out to 10^51, only 9^51 numbers don’t have a 4 in them, and that’s about 4.6 * 10^48, or less than one half of one percent, which means a bit more than 99 and a half percent include a 4. Rounds up to 100%.