A person wrote down all numbers from 1 to 999 inclusive. How many times was the digit 8 written? I'm not sure how to approach this one... Should I group these into 100's and add up... Or is there a more simple approach to this? Thanks!
P.S: Wasn't sure what the tag was for this one. Sorry.
On
Yes, group them into groups of a hunderd and add up. All of them have the seame number of digits $8$, apart from the $800$-$899$ group, which has a hundred more eights than the other groups.
So, how many are in a group of a hundred? Split it into groups of ten, and you see that the $80$-$89$ has eleven, while the other groups have one each. So that's $20$ digit $8$'s in each group of a hundred. Can you add them together now?
On
We have three cases: numbers that have three $8$, number that have two $8$ and numbers that have one $8$.
There's only one number with three $8$: $888$.
With two $8$, we have _ _ _ three spots to put them, thus $\binom{3}{2} = 3$ options. After that, we have to choose between 9 items ($0$ through $9$, but not the $8$) to put in the remaining place (we're covering two-digit numbers -i.e, the 88- when we choose $0$ for the first spot). So there are $3 \cdot 9 = 27$ of these.
With only one eight, we have to choose its place from three options _ _ _, then choose the other two numbers with $9$ options in each step (again $0$ through $9$, but not $8$), so $3 \cdot 9 \cdot 9 = 243$ numbers.
So we wrote the $8$ exactly $1\cdot3 + 27\cdot2 + 243\cdot1 = 300$ times.
Start with all the numbers that end in 8.
Their digits look like _ _ 8, and there are 10 choices for each of the other digits. Hence, there are 100 8s in the one's place.
Similarly, for the 10s place: _ 8 _, there are 10 choices for each again, and there are 100 8s in the 10s place.
Finally, using the same technique there are again 100 8s in the 100s place, for a total of 300 8s written down.
Note that this will work for any digit except 0, since we (usually) don't write down leading 0s.