How many integers between 1 and 10^21

E

For the combinations of 2's and 0's we have one 2 and 20 0's

total combinations =\(\frac{21!}{20!}\)= 21 -> because 20 0's are identical.

For the combinations of 1's and 0'2 we have two 1's and 19 0's

Total combinations = \(\frac{21!}{19!*2!}\)= 210

Total = 210+21 = 231

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

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between 1 and 10^21 means we are talking about numbers with upto 21 digits.

Case 1
Numbers of the form 2,20,200,etc
Possible numbers = 21 (1 possibility for each number of digits)

Case 2
Numbers using the digits {1,1}
Such a number will have a minimum of 2 digits and a maximum of 21 digits, with the first digit=1.
Now, consider the k-digit number, with the first digit "1", out of the rest of the k-1 digits, we have k-1 ways to choose where to put the second 1. So k-1, k-digit numbers possible.
So total number of numbers = Summation(k-1;k=2 to k=21)=1+2+...+20=20(21)/2=210



So total number of numbers = 210+21=231

Answer is (e)
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so the second condition means there are 21 positions and two number, 1 and 1 so place in then. hence we can use the combination formula 21c11 and get 210.

Yea that's a good way to think about it

But formula will be C(21,2) not C(21,11). C(21,2)=210

Posted from my mobile device
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Wow!! whenever I see these kinda questions I got stunned but guys like bunuel,shrouded1, and gurpreet are there to help . Nice method gurpreet and whenever a quant question is posted, shrouded1 is always there to explain it. Thank you! guys (I couldn't solve it but now i know how to solve it )

Nice question..... 21C2 + 21 = 210 + 21 = 231

Mike and Macfauz

are'nt there 22 places to choose for placing two 1s??

\(10^{21}\) has 22 places. But we are asked for "between 1 & 10^21". The greatest integer less than 10^21 will have only 21 places.

Combinations of digits that sum up to 2 are either 1+1 or 2+0:

Combinations 1+1:

Double Digits
11 Total: 1

Triple Digits
101
110 Total: 2

Quadruple Digits
1001
1010
1100 Total: 3

So now the pattern emerges that for every N digit number, there are N-1 combinations summing up to two. The number of integers should be between 1 and 10^21, exclusive which means that the largest allowed integer does only have 21 digits, not 22 which means that the amount combinations with the largest amount of places is 20 (21-1).

We can thus calculate the total amount of integers that satisfy the question by calculating the set of consecutive integers: 1+2+3...+20 which is 210. However, we did not account for the combinations of 2 and 0 that sum up to two, we have to add them first:

Combinations 2+0:

Single Digit
2 Total: 1
Double Digits
20 Total: 1
Triple Digits
200 Total: 1

This pattern is obviously easier to comprehend and will account for an additional 21 combinations.

Hence: 210+21 = 232.

Sorry for my rusty English, I am not a native speaker .

Hello all. This is my first question. Hopefully someone could clarify.

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

My answer: : Every level has 2 digits when there are two 1's in the number.

Level 1 : 11 = 1
Level 2 : 101 & 110 = 2
Level 3 : 1001 , 1010 & 1100 = 3
Level 4 : 10001, 10010, 10100 & 11000 = 4

So apparently every level you rise, 1 more integer with sum of digits 2 is added.

In this case,starting from the first "0" this number -> 1.000.000.000.000.000.000.000 (=10^21) has 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21 =231 which is exactly the correct answer.

However, the OFFICIAL answer mentions this solution:

10^21 is a 22 digit number, this means that all integers betweeen 1 and 10^21 will have at most 21 digits. Now for the purpose of this question we will say that all the numbers in this range have exactly 21 digits. We do this by including several leading 0's of a number. So e.g. we can write 10 million 1 hundred thousand as 000.000.000.000.010.100.000 Similarly we could write 20.000 as a 21 digit number. Now for a number to have the sum of its digits to equal 2 we need to consider 2 possible cases. Case 1 is that we have 2 1's in our number and then 19 0's. Case two is that we have one 2 and 19 zero's. To find out the number of 21 digits number with 19 zero's and two 1's. This we can do in 21C2 ways. If we plug this in the nCr formula we get 210. In Case two we have 21C1 = 21 ways. So 210 + 21 = 231.

Is my approach wrong?? I think the "official" answer is quite strange since 000.020.000.000.000.000.000 is not really an integer?

Any help is welcome, thanks in advance!!

Merging similar topics. Please refer to the solutions above.
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[quote="JeroenReunis"]Hello all. This is my first question. Hopefully someone could clarify.

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

My answer: : Every level has 2 digits when there are two 1's in the number.

Level 1 : 11 = 1
Level 2 : 101 & 110 = 2
Level 3 : 1001 , 1010 & 1100 = 3
Level 4 : 10001, 10010, 10100 & 11000 = 4

So apparently every level you rise, 1 more integer with sum of digits 2 is added.

In this case,starting from the first "0" this number -> 1.000.000.000.000.000.000.000 (=10^21) has 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21 =231 which is exactly the correct answer.

However, the OFFICIAL answer mentions this solution:

10^21 is a 22 digit number, this means that all integers betweeen 1 and 10^21 will have at most 21 digits. Now for the purpose of this question we will say that all the numbers in this range have exactly 21 digits. We do this by including several leading 0's of a number. So e.g. we can write 10 million 1 hundred thousand as 000.000.000.000.010.100.000 Similarly we could write 20.000 as a 21 digit number. Now for a number to have the sum of its digits to equal 2 we need to consider 2 possible cases. Case 1 is that we have 2 1's in our number and then 19 0's. Case two is that we have one 2 and 19 zero's. To find out the number of 21 digits number with 19 zero's and two 1's. This we can do in 21C2 ways. If we plug this in the nCr formula we get 210. In Case two we have 21C1 = 21 ways. So 210 + 21 = 231.

Is my approach wrong?? I think the "official" answer is quite strange since 000.020.000.000.000.000.000 is not really an integer?

Any help is welcome, thanks in advance!![/quote

hi Jeroen ...your approach is ok but u r going wrong on two accounts although u r getting the correct ans
1) as far as your counting is considered for two 1's...
the last part where u have considered 21 at level 21 is wrong.. in ur way there should be only 20 levels..
the reason is 10^21 is the lowest 21 level digit number..that is 10000....000. and one lower than it is 999..20 times so it cannot have 1000..20 times...01
2) so the ans is 1+2+...+20, which is 210.. now what about rest 21.. this is the second mistake.. what about digit 2 with all rest digits as 0 for eg 2,20,200..
2*10^0,2*10^1,2*10^2,.. till 2*10^20=21 such numbers
combine the two 210+21=231 s
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The answer is exactly correct but you got lucky!

You will go till Level 20, not level 21 because level 20 will have all 21 digit numbers. Level 21 has 22 digit numbers but 22 digit numbers will be greater than 10^21 because 10^21 is the smallest 22 digit number.
Then how come you get the correct answer? Because you have to add 21 to the sum for these 21 numbers:
2
20
200
2000
20000
200000
....
There will be 21 such numbers. The sum of digits here too is 2.




THink about this: Is 020 an integer? Yes. It is 20. (0 on the left of an integer has no value.)
If we think this way, we don't have to consider the numbers with different number of digits separately. We can say that all number have 21 digits such that 0 can be placed in the beginning too.
00000...0000000002
00000...0000000020
00000...0000000200
00000...0000002000
...
The same numbers as given above can be written like this.

Makes sense?
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I think the options to the above questions fall short of the correct answer

Here is my solution-



A- 2-------10^0 series-----------------1 (0+1)
B- 11,20------------10^1 series----------------------2 (1+1)
C-101,110,200---------10^2 series----------------------------3 (2+1)
D-1001,1010,1100,2000-------10^3 series--------------------------4 (3+1)
E-10001,10010,10100,11000,20000--------10^4 series-------------------------5 (4+1)

and so on......................................................10^20 series..............................................21 (20+1)


so essentially we have to find the sum of the series

1+2+3+................................+.20+21 = 231

Experts ..can u validate the above
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hi sami,
u have done everything perfect except one point...
22 which u have added in10^21 is not to be added..
reason 10^21 is the only number and the lowest one in that series.. so u have to add 0 for that series..
ur ans will become 253-22=231..
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Thanks!! i just realised the mistake ....
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If the sum of the digits of an integer is 2, then either a digit is 2 and the rest of the digits are 0 OR two digits are 1 and the rest of the digits are 0.

Case 1: One digit is 2 and the rest of the digits are 0.

Since the first digit of an integer can’t be 0, 2 must be the first digit. So between 1 and 10^21, we have: 2, 20, 200, 2000, etc. Since 2 = 2 x 10^0, 20 = 2 x 10^1, 200 = 2 x 10^2, …, and the last one being 2 x 10^20, we have 21 such numbers.

Case 2: Two digits are 1 and the rest of the digits are 0.

Notice that 10^21 is 1 followed by 21 zeros, i.e., a 22-digit number. However, it’s the smallest 22-digit number. Therefore, any number we have can only have at most 21 digits. Also, though we’ve mentioned that the first digit of an integer can’t be 0, let’s for the moment assume that it can be. Now assume that we have a 21-digit number (or code) that consists of 2 ones and 19 zeros. The number of ways to arrange 2 ones and 19 zeros is 21!/(2! x 19!) = (21 x 20 x 19!)/(2 x 19!) = 210. So there are 210 such codes and all these 210 codes are actually the numbers from 1 to 10^21 that have two digits of 1 and the rest of the digits as 0 if we omit all the initial 0s before the first digit of 1 (for example, consider 000000000000000000101 as the number 101).

Therefore, in the two cases combined, we have a total of 21 + 210 = 231 such numbers.

Answer: E
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Provided that you adjust the approach according to the corrections illustrated by Bunuel and others after this message, this is the best approach in my opinion.
Especially if you aim for 49-50 where the ROI on each hour invested on Q preparation is still human. (personal and contestable opinion)