integer whose digits add up to 2

[kris610]

How many integers between 1 and 10^21 are such that the sum of their digits is 2?

A 190

B 210

C 211

D 230

E 231

E

Best approach to solve this in < 2 mins?

[Night reader]

10^21 is 22-digit number we can have digit "1" at the beginning of each starting number than accordingly digit "1" afterwards - in trillions, millions, tens
In 22-digit number we can have 22 "1"s //likewise in 100 we can have 110 101, we deduce only 2 numbers possible to have digits summed up to 2// hence here the number of integers is 21
next we have 21-digit number --> 20
19
18
17
16
15
14
13
12
11
10
...
we notice sequence (21,20,... 1) the sum of sequence is [(1+21)/2]*21=231

IOM E

i spent 60 sec. really

How many integers between 1 and 10^21 are such that the sum of their digits is 2?

A 190

B 210

C 211

D 230

E 231

E

Best approach to solve this in < 2 mins?

[Anurag@Gurome]

How many integers between 1 and 10^21 are such that the sum of their digits is 2?

A 190
B 210
C 211
D 230
E 231

The integer must be composed of any one of the following formats,

• 1. Two 1's and rest zero
  2. One 2 and rest zero

And the integer can have at most 21 digits.

Therefore, number of such integers = (Number of ways to put one 2 in any of the 21 places) + (Number of ways to put two 1's in any two of the 21 places)

Now, number of ways to put one 2 in any of the 21 places = 21
And, number of ways to put two 1's in any two of the 21 places = 21*20/2 = 210

(This is because first 1 can be placed at any of the 21 positions and second 1 can be placed at any of the remaining 20 positions. Now there will be a copy of each of the combination as two 1's are identical. Hence divide by 2)

Therefore, number of such integers = 21 + 210 = 231

The correct answer is E.