Greenlight Instructor,
Sir can you please post similar type of question.

This makes no sense to me. Oh well.

This is a very confusion question, I'm not sure this would even appear on the exam, though maybe an expert could correct me if I'm wrong.

Initially, I thought the question was asking: "what numbers does 1 appear in, from 1-300."

That is not what the question is asking.

The question is asking, if we were to write all the numbers from 1-300, how many times would we write the number 1?

So for example, when we write the number 10, there is one occurrence of the number 1. In 11, there are two occurrences of 1, and in 111 there are three occurrences of 1.

How to solve this is a little complicated too, but it boils do counting principles.

Create three placeholders:

_ _ _

And from here you start expressing the numbers 1-300.

1 can be written like this:

0 0 1

10 can be written like this:

0 1 0

And so on. It's about noticing we need to count the number of occurrences in each unit, tens, and hundreds position.

Let's start with 1-100, and the rest should fall into place with some logic.

1-100

1 occurs a total of 10 times between 1 and 100 in the units digit:

0 0 1
0 1 1
0 2 1
.
.
.

1 occurs a total of 10 times between 1 and 100 in the tens digit:

0 1 0
0 1 1
0 1 2
.
.
.

Now we can notice that this pattern would be the same for 101-200 and 201-300, so we will repeat the ones digit pattern three times, and the tens digit pattern three times.

10*3 + 10*3 = 60.

Now that we covered the units and tens digit, we can't forget about the hundreds digit. Only the numbers 100-199 have the number 1 in the hundreds digit with the range given. So that's 199-100+1 = 100.

So in total we have 100 + 60 = 160.

Giving 160 as the answer.

Can you share your solution please? I tried using P & C only but couldn't get 160.

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