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

Question

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

chetan2u · Accepted Answer

Let us calculate different digit number possible:-
1) Single digit - Only one number, that is 2
2) 2-digit - 2A or A2.....1*10+9*1 = 19, subtract the common in A2 and 2A, which is 22.  so, 19-1 =18 
When 2 is tens digit, A can take any of the 10 digits as value, and when 2 is units digit, A can take any of 9 digits.
Total = 1+18 or 19 single and 2-digit numbers.
3) 3-digit numbers from 100 to 199
19 as above

Total 19+19 = 38.

Now from 200 to 299, all integers will contain 2, but we are looking at 100-38 or 62nd number.
So, 200+62-1 or 261 (  we subtract 1 as 200 in inclusive.)

A

KarishmaB · Answer

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)

Archit3110 · Answer

from 0 to 99
total digits with 2 in list are 19
100 to 199; again 19
38 listed numbers 
100-38 ; 62
200 is 39th number so 61st number is 100th number
261
option A

AnkurGMAT20 · Answer

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

Bunuel · Answer

Added the tag. Thank you!

snow12snow · Answer

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

sayan640 · Answer

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

gmatophobia · Answer

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 Screenshot 2024-03-30 115428.png Step 2 : Now, let's find out the number of integers between 100 and 199 that includes a 2. We have 19 such numbers Screenshot 2024-03-30 115530.png 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 Screenshot 2024-03-30 115710.png 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

DmitryFarberMPrep · Answer

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.

rajpatil.june · Answer

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!

arnavghatage · Answer

Bunuel 
Please can you suggest where I can find similar questions?

Bunuel · Answer

msignatius · Answer

The calculation can be made quicker if you realize:

0-9 is 10 numbers.
0-99 is 100 numbers.

So, clearly, if we're looking for each tens digit 0 with an accompanying units digit 2, we get 10 such numbers.

Also, for the 20-29 range, we too get 10 numbers with 2, but we've already accounted for the 2 above, so that's 10 + 9 = 19.

For up to 199, add another 19. The math doesn't change. 38 it is now.

We're left with another 100 - 38 = 62 numbers left. And we know, somewhere under 200, was the 38th integer with 2 in it. Every integer starting from 200 will contain a 2.

Now, again, remember, 0-9 is 10 numbers. 0-99 is 100 numbers. So, if we want to find the 100 - 38 = 62nd number in the list starting from 200, we're going up to 261.

261. That's your answer.

Frost777 · Answer

Hi,If we are counting 22, then wont we also have to count 220-229?

Bunuel · Answer

Yes, 220–229 are counted.

The point is that we are counting integers, not occurrences of the digit 2. So 22 is counted once, not twice, and 222 is also counted once, not three times.

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

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

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