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by shashank.ism » Tue Feb 09, 2010 7:18 am
Suppose n is an integer such that the sum of the digits of n is 2, and 10^10 < n < 10^11. The number of different values for n is

11
10
9
8
7
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by ajith » Tue Feb 09, 2010 11:10 am
shashank.ism wrote:Suppose n is an integer such that the sum of the digits of n is 2, and 10^10 < n < 10^11. The number of different values for n is

11
10
9
8
7
there is only 1 starting with 2 and rest 10 zeros
Starting with 1 there are 10 of them (11--9 zeros 101 --8zeros....)

In total 11 of them
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by shashank.ism » Tue Feb 09, 2010 11:31 am
ajith wrote:
shashank.ism wrote:Suppose n is an integer such that the sum of the digits of n is 2, and 10^10 < n < 10^11. The number of different values for n is

11
10
9
8
7
there is only 1 starting with 2 and rest 10 zeros
Starting with 1 there are 10 of them (11--9 zeros 101 --8zeros....)

In total 11 of them
Hmm nice apporach to the problem...
If the sum of no. is like 7 or 8 then what we will do is there any methodical peocedure....
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by harsh.champ » Tue Feb 09, 2010 12:08 pm
shashank.ism wrote:
ajith wrote:
shashank.ism wrote:Suppose n is an integer such that the sum of the digits of n is 2, and 10^10 < n < 10^11. The number of different values for n is

11
10
9
8
7
there is only 1 starting with 2 and rest 10 zeros
Starting with 1 there are 10 of them (11--9 zeros 101 --8zeros....)

In total 11 of them
Hmm nice apporach to the problem...
If the sum of no. is like 7 or 8 then what we will do is there any methodical peocedure....
10^10 < n < 10 x 10^10
Now,if the sum is 2
Case 1 -:2 1's are there
10^10 means 10 zeros
so we have 11 places.
Selection of 1 place can be done in 10 ways.
Case 2:- 2 x 10^10

Hence ,11 A
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