2016_Question_1

2016_Question_1

There is probably a neater way but here we go!

So if we count the number of times the digits 1, 3, 5, 7, and 9 appear in the list we are done.

To simplify things a bit, we will first only count in the list 1, 2, 3, …1999

1’s
So in the interval 1 - 99 we have 10 occurrences of 1 in the one’s digit and 10 occurrences of 1 in the ten’s digit. Thus we have twenty ones.

In the interval 100 -199 we have the twenty occurrences plus one hundred more since every hundred’s digit will be one.

In total, we have 300 occurrences of 1 in the interval 0 - 999 and we can generalize this for all the odd digits.

The same applies to the interval 1000-1999 except that the 1 is repeated an additional 1000 more times.

Thus our total count is 600 \cdot 5 + 1000=4000.

Now let’s count the number of occurrences from 2000 - 2016.
Quick counting gives us 15 but we could also use the list 1, 2, 3, … 11 and just add 7 for (1)2, (1)(3), (1)4, (1)(5), (1)6.

The answer is 4015.

QED.

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