Problem Solving

How do we do these kind of problems?


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

OA - 231

singhsa wrote:How do we do these kind of problems?


How many integers between 1 and 1021 are such that the sum of their digits is 2?

OA - 231

We have the biggest possibility as a 4-digit integer 1010, so let's deal with the smallest possibility as 0002 only; and for 4-digit integers, 3 separations (say $$$) are required. Let's represent the sum of 2 by two asterisks **, and as we've allowed ourselves to take '0' into account in order to complete the four digit-locations, more than one separations can be put at a given digit-location.

With this, 0002 or 0011 may now be taken as $$$** or $$*$*, respectively. In other words, the question in chief is asking us to find all the permutations for the five locations like $$$** or $$*$*. It has got 2 items of one type and 3 of another type, total is 5, and the permutations are

= 5!/ (2! 3!)

= 10

We can count this on finger tips too

0002
0020
0200
0011
0101
0110
1001
1010

Since we cannot include 1100 and 2000, hence [spoiler]8 only.


I don't know why and how it's 231.[/spoiler]

Thanks sanju09 for sharing the technique, I am sure it would help for the bigger summations too, where we cannot count the possibilities on the tip of fingers. There is something wrong either with the question or with the oa, please check it singhsa.

sanju09 wrote:

singhsa wrote:How do we do these kind of problems?


How many integers between 1 and 1021 are such that the sum of their digits is 2?

OA - 231

We have the biggest possibility as a 4-digit integer 1010, so let's deal with the smallest possibility as 0002 only; and for 4-digit integers, 3 separations (say $$$) are required. Let's represent the sum of 2 by two asterisks **, and as we've allowed ourselves to take '0' into account in order to complete the four digit-locations, more than one separations can be put at a given digit-location.

With this, 0002 or 0011 may now be taken as $$$** or $$*$*, respectively. In other words, the question in chief is asking us to find all the permutations for the five locations like $$$** or $$*$*. It has got 2 items of one type and 3 of another type, total is 5, and the permutations are

= 5!/ (2! 3!)

= 10

We can count this on finger tips too

0002
0020
0200
0011
0101
0110
1001
1010

Since we cannot include 1100 and 2000, hence [spoiler]8 only.


I don't know why and how it's 231.[/spoiler]

I'm really sorry guys....its 10^21 and not 1021.

Extremely sorry for that.

With the revised question I get 253

The numbers can be in 2 forms

(a) 2 and a combination of 0s from 1 to a 10^21 (a 22 digit number) which sum to 2 ----> a total of 22 such numbers can be set up with 2 occupying each of the 22 digits once and 0s everywhere else

(b) a combination of 1s from 1 to a 10^21 (a 22 digit number) which sum to 2 ----> a total of 231 such numbers are possible
starting from 11; a pattern quickly emerges for 2 digit numbers only 1 such number is possible; for 3 digits only 2 such numbers are possible (101 and 110); for 4 digits only 3 such numbers are possible...... for 22 digits only 21 such numbers would be possible...

so 1+2+3.....+21 = 231

So it should be 231 + 22 = 253

How about 20....Is this a 2 digit integer with sum of 2 digits =2. Your following statement says from 1 to 99 only 1 integer is possible i.e. 11. I think 20 is another number in 1 to 99 where sum of integer is 2.

B

How many integers between 1 and 1021 are such that the sum of their digits is 2?

between 1 and 10 - one number i.e. 2
between 11 and 20 - two numbers i.e. 11 and 20
between 21 and 30 - one number i.e. 29 (careful sum of digits is 2)
and so on...

Between 1 and 100 - 11 numbers
between 101 and 200 - 11 numbers
.
.
.
.
between 2001 and 2100 - 11 numbers

answer = 20*11 = 220

for answer to be 231, number range must be from 1 to 2200.
Please check.