If all positive integers that have at least 1 digit equal to 2 are lis

Bunuel chetan2u this is an FE official question. Please tag it.

Added the tag. Thank you!

why does no one count 220-229? it adds 10 more numbers, so the final answer should be 251 not 261

gmatophobia , Can you please provide an easy explanation to this ?

 

sayan640 We can solve this question by counting the integers between a given range -

Step 1: Let's find out the number of integers between 1 and 99, both inclusive that consists of a 2 

We have 19 such numbers



Step 2: Now, let's find out the number of integers between 100 and 199 that includes a 2.

We have 19 such numbers


 ­
Step 3: We have obtained 38 integers so far, so let's find the number of integers in the next block of 200 to 299 that includes a 2.

All the numbers include at least one 2. Hence the number of integers = 10 * 10 = 100


 
Step 4: We now have 100 + 38 = 138 integers, however we need only 100 integers. So we have to let go of 38 integers from the last obtained integer, 299

299 - 38 = 261

Option A­

snow12snow
We're not concerned with how many total 2's we see, just with how many INTEGERS we see that have 2 in them. So whether our number is 2, 22, or 222, it still just counts as one case.

It may also help to note that the numbers from 101 to 199 will work just like the numbers from 1 to 99, since it's just the same list with a 1 in front. So once we establish that we have 19 in the first set (1 for each "decade," except for the 20's, where all 10 contain a 2), then we know that the next set is the same. So from 1-199, there are 38 integers containing a 2. We need 62 more. However, we now hit 200, at which point EVERY number contains a 2. So we just need to count the 62nd number in the 200's. Counting from 201 up to 262 would be 62 numbers, so from 200 we just need to count up to 261.