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by knight247 » Sun Jul 17, 2011 11:41 pm
How many integers are there between 2*10^5 and 9*10^5 so that the sum of their digits is 3?
(A)9
(B)8
(C)7
(D)6
(E)5

Answer is D. Solved this without using combinatorix. Combinatorix being my weakness i'm hoping someone can solve this using combinatorix and clearly explain the steps. Thanks
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by Geva@EconomistGMAT » Sun Jul 17, 2011 11:52 pm
knight247 wrote:How many integers are there between 2*10^5 and 9*10^5 so that the sum of their digits is 3?
(A)9
(B)8
(C)7
(D)6
(E)5

Answer is D. Solved this without using combinatorix. Combinatorix being my weakness i'm hoping someone can solve this using combinatorix and clearly explain the steps. Thanks
Forget combination theory - it would just confuse in this case. For these small answer choices, it's best to simply count.

We start with 2*10^5 = 200000. We already have a 2, so to reach the sum of digits of 3, we just need to add 1. That 1 can come replace one of the 5 zeroes:
200001
200010
200100
201000
210000

5 options.

Next, move to the 300000 - and stop there. Already we have a sum of digits of 3: 3+0+0+0+0+0. We can't add any other digit, or we get a sum of digits greater than 3.

The same can be said for any number beginning with 4, or 5, or 6 - they're already greater than the sum of digits of 3, so they're out. So there are 6 options altogether - five options beginning with 2, and one option beginning with 3.
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by abhisays » Sun Jul 17, 2011 11:54 pm
Only these six possibilities are there.

200001
200010
200100
201000
210000
300000

Hence answer should be D
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by Anurag@Gurome » Sun Jul 17, 2011 11:59 pm
knight247 wrote:How many integers are there between 2*10^5 and 9*10^5 so that the sum of their digits is 3?
The question is asking how many 6 digit integers are there between 200,000 and 900,000 such that sum of their digits is 3. Now the integer can be made of one 2, one 1 and four zeroes OR one 3 and rest zeroes.

For the first case, first digit is occupied by 2. And 1 can take any of the 5 remaining places.
So number of such integers is 5.

And for the second case only one integer is possible, i.e. 300,000

Hence total number of such integers is = (5 + 1) = 6

The correct answer is D.
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by amit2k9 » Mon Jul 18, 2011 7:04 am
possible combinations are 2,1 and 3,0.

5 positions for 1 hence 2,1 = 5
1 position for 3 hence 3,0 = 1

thus 6 in total.
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