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# How many integers between 1 and 10^21 are such that the sum

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How many integers between 1 and 10^21 are such that the sum [#permalink]  04 Dec 2012, 14:47
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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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Mike McGarry
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Re: the sum of the digits is 2 [#permalink]  04 Dec 2012, 21:39
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]  05 Dec 2012, 02:55
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]  05 Dec 2012, 03:13
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]  28 Dec 2012, 10:10
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

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 .
Re: How many integers between 1 and 10^21 are such that the sum   [#permalink] 28 Dec 2012, 10:10
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