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

Added the tag. Thank you!
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why does no one count 220-229? it adds 10 more numbers, so the final answer should be 251 not 261
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gmatophobia , Can you please provide an easy explanation to this ?
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sayan640
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

Attachment:
Screenshot 2024-03-30 115428.png
Screenshot 2024-03-30 115428.png [ 53.47 KiB | Viewed 5214 times ]

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

We have 19 such numbers

Attachment:
Screenshot 2024-03-30 115530.png
Screenshot 2024-03-30 115530.png [ 55.21 KiB | Viewed 5180 times ]
 ­
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

Attachment:
Screenshot 2024-03-30 115710.png
Screenshot 2024-03-30 115710.png [ 56.85 KiB | Viewed 5141 times ]
 
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
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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.
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raheelsn
If all positive integers that have at least 1 digit equal to 2 are listed in increasing order, what is the 100th integer on the list?

A) 261
B) 262
C) 270
D) 271
E) 279
Very important to keep in mind for "number of times a digit appears" questions:­
In consecutive 100 numbers, we have 19 integers with a certain digit (any from 1 to 9) in units or tens (or both) place. For example in 1 to 100, 5 appears in 19 integers. It appears 20 times because it appears twice in 55 but it appears in 19 integers.  

So from 1 to 199, we have 19*2 = 38 integers with 2 in them. 
Starting from 200, every integer has 2 in it. We need 62 more integers so 261 will be the 100th such integer.

Answer (A)
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Consider range 0-00 first,

numbers having 2 as a digit at units place are 2, 12, 22, 32 .... 92. So total 10
Similarly numbers having 2 as a digit at tens place are 20, 21, 22..... 29 So total 10

22 is counted in both the cases so total will be 10+10-1 = 19

Similarly, in range 100-199, we have 19. Total 38.

Now in range 200-299 all numbers satisfy the range, we need 100th numbers. 100 -38 = 62.
starting 200 62th will be 261.

So A.

Thanks!
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