How few 4’s?
May 9, 2006 am31 4:02 am
What percent of numbers (ie, counting numbers, 1, 2, 3, 4, … and keep going) contain the digit '4'?
Round your answer to the nearest whole percent.
(I will look for my source, but I think this has been around a long time)
12 Comments
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JD2718 
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?
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.
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.
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.
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?
Um, never mind.
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*
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.
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. : )
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?
I originally found this puzzle at The Grey Labyrinth, a puzzles site
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%.