Bruno Pittman

2022-07-01

Let's call $S$ the infinite string that is made by concatenating the consecutive positive integers written down in base $10$. Thus,
$S=12345678910111213141516171819202122232425\dots$
Any number in $S$ occurs multiple times. The first occurrence of $3$ is in the third position of the series, the second occurrence is in the seventeenth position, and so on.
How do I find the position of the hundredth occurrence of $3$? Is there a pattern?

Elianna Wilkinson

Expert

You can solve this by thinking of some lists:
How many 3's is there in 1-10? -- 1
How many 3's is there in 11-20? -- 1
How many 3's is there in 21-30? -- 2
How many 3's is there in 31-40? -- 10
How many 3's is there in 41-50? -- 1
...
How many 3's is there in 91-100? -- 1
Now, how many 3's is there in 1-100?
How many 3's is there in 101-200?
How many 3's is there in 201-300?
How many 3's is there in 301-400? (obviously sufficient)
Then, when you know the what number comes at the 100:th place you just need count the place - for example by "counting lists" again.

skynugurq7

Expert

Try some careful counting. If mine is careful enough, I make the 100th occurence of 3 to be in the 889th position.
It's the first digit of 333.
Counted like this:
In 1 to 99 there are 10 threes in the unit's digit and 10 in the ten's digit, so that's makes 20, and hence there are 3*20=60 in 1 to 299.
Now in 300 to 329 there are 30 in the hundred's digit and 3 in the unit's digit, so that's another 33, making 93 so far. Now just look at 330331332333 and you will see that the 100th three is the first digit of 333 which, by some straightforward calculations, is in the 889th position.

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