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by ace_gre » Wed Jan 06, 2010 3:21 pm
Hi, Here is my approach. We need to find the sum of digits equal to 5 for all numbers upto 9999.
Unique combinations are as follows:

1) 5,0,0,0
2) 4,1,0,0
3) 3,2,0,0
4) 3,1,1,0
5) 2,2,1,0
6) 1,1,1,2

Since the problem asks for all numbers less than 9999, we need to include single,double,three and four digit numbers.
Now it becomes a permutation problem.

1) Thousand place can be filled with 4 digits, hundreds with any of the remaining 3 and so on. And since there are three zeros, there are duplications. So divide by 3!. i.e. 4!/3! = 4

2) 4!/2! = 12

3) 4!/2! = 12

4) 4!/2! = 12

5) 4!/2! = 12

6) 4!/3! = 4.

Adding all the above IMO, C-> 56.
Please post the OA.
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by onedayi'll » Wed Jan 06, 2010 8:45 pm
Agree with 'ace_gre'

Answer - C
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by rdchandvadkar » Wed Jan 06, 2010 10:07 pm
thanks for the responses..

the correct answer is C - 56

but could you pls explain the permutations part of the problem a lil more in detail. my permutation and combination section is the weakest.

thanks.
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by ace_gre » Thu Jan 07, 2010 10:58 am
Please check this site for more information..

https://www.tutors4you.com/permutationco ... torial.htm

Let's take 2,3,0,0 combination. Here the possible numbers where sum = 5 are
23, 32, 230,203,320,302,2300,2003,2030,3200,3002,3020.

Here all numbers less than or equal to 9999 are asked. So the problem becomes how to arrange 2,3,0,0.
Here 0 can still be the thousand digit and hundred digit ( true in case of 23 and 32)..

ABCD ==>No. of ways of filling digit A is 4(can be 2,3,0,0)=4
No. of ways of filling digit B is 3 (remaining numbers after A is chosen)
No. of ways of filling digit C is 2(remaining numbers after A and B are chosen)
No. of ways of filling digit D is 1(last number remaining)

Total number = 4*3*2*1/2. (Divide by 2 since there are two zeros. Using any of the two will result in same number)
==>12.

Same approach for other combinations as well.

Hope its clear. :)
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by rdchandvadkar » Sat Jan 09, 2010 4:15 am
thanks.. absolutely clear now.. thanks again for the help...
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by linkinpark » Sat Jan 09, 2010 8:09 am
ace_gre your solution is understood but i was thinking otherway round but got stuck, tell me what do you think

maximum num = 9999, a 4 digit number = 1000(x) + 100(y) + 10(w) + z

x+y+z+w = 5 is what we want, for digits 0 to 5 we can arrange to get sum as 5 which we can do in 5!/2 ways right? and here I'm stuck coz there is no 60 answer :(
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by ace_gre » Sat Jan 09, 2010 4:23 pm
for digits 0 to 5 we can arrange to get sum as 5 which we can do in 5!/2 ways right?
linkinpark, I am sorry cannot seem to understand how you arrived at this :?
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by Brent@GMATPrepNow » Sat Jan 09, 2010 4:49 pm
rdchandvadkar wrote:How many positive integers less than 10,000 are there in which the sum of the digits equals 5?

a. 31
b. 51
c. 56
d. 62
e. 93

This is a very hard question that I created for ReadyForGMAT.com
As it says on the document, this question is beyond the scope of the GMAT, but here's a quick solution nonetheless


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