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How many integers between 1 and 10^21

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How many integers between 1 and 10^21 [#permalink] New post 15 Oct 2010, 23:50
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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
[Reveal] Spoiler: OA

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Re: How many integers between 1 and 10^21 [#permalink] New post 15 Oct 2010, 23:55
Pkit wrote:
How many integers between 1 and \(10^21\) (10 in 21st power) are such that the sum of their digits is 2?

a.190
b.210
c.211
d.230
e.231

Bunuel, please leave a chance for other people to solve this. :)

thanks


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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Re: How many integers between 1 and 10^21 [#permalink] New post 16 Oct 2010, 02:15
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.
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Re: How many integers between 1 and 10^21 [#permalink] New post 16 Oct 2010, 02:25
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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

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Re: How many integers between 1 and 10^21 [#permalink] New post 16 Oct 2010, 02:49
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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
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Re: How many integers between 1 and 10^21 [#permalink] New post 16 Oct 2010, 04:33
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 :) )
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How many integers between 1 and 10^21 are such that the sum [#permalink] New post 04 Dec 2012, 13:47
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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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Re: the sum of the digits is 2 [#permalink] New post 04 Dec 2012, 20:39
Nice question..... 21C2 + 21 = 210 + 21 = 231
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Re: How many integers between 1 and 10^21 are such that the sum [#permalink] New post 05 Dec 2012, 01:55
Mike and Macfauz

are'nt there 22 places to choose for placing two 1s??
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Re: How many integers between 1 and 10^21 are such that the sum [#permalink] New post 05 Dec 2012, 02:13
ratinarace wrote:
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.
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Re: How many integers between 1 and 10^21 are such that the sum [#permalink] New post 28 Dec 2012, 09:10
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 :).
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Hard counting problem? New way to solve?? [#permalink] New post 29 Jan 2015, 10:52
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!!
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Re: How many integers between 1 and 10^21 [#permalink] New post 29 Jan 2015, 10:54
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JeroenReunis wrote:
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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Re: How many integers between 1 and 10^21 [#permalink] New post 29 Jan 2015, 18:56
[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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Re: How many integers between 1 and 10^21 [#permalink] New post 29 Jan 2015, 22:12
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JeroenReunis wrote:
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.



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.


JeroenReunis wrote:
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!!


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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How many integers between 1 and 10^21 [#permalink] New post 29 Jan 2015, 23:20
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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Last edited by samichange on 29 Jan 2015, 23:40, edited 1 time in total.
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Re: How many integers between 1 and 10^21 [#permalink] New post 29 Jan 2015, 23:30
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samichange wrote:
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......................................................10^21 series..............................................22 (21+1)


so essentially we have to find the sum of the series

1+2+3+................................+.21+22 = 253

Experts ..can u validate the above



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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Re: How many integers between 1 and 10^21 [#permalink] New post 29 Jan 2015, 23:38
chetan2u wrote:
samichange wrote:
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......................................................10^21 series..............................................22 (21+1)


so essentially we have to find the sum of the series

1+2+3+................................+.21+22 = 253

Experts ..can u validate the above



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..



Thanks!! i just realised the mistake ....
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Re: How many integers between 1 and 10^21   [#permalink] 29 Jan 2015, 23:38
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