Problem Solving

phoenixhazard wrote:

GMATGuruNY wrote:

skalevar wrote:How many positive integers are less than 10,000 in which the sum of the digits is equals 5?

A. 31
B. 51
C. 56
D. 62
E. 93

I got started on this, but could only try to count up the ways manually. There must be a trick. Please help!

OA. is C. From Ready for GMAT.com Advanced Quant download.

Here is an efficient way to solve this problem.

Let's say that we have the 5 letters DDDDD.

However we separate the 5 letters, we will still have 5 letters:

DD DD D
D DDD D
DDDD D
DD DDD

We can think of these various groups of D's as digits that add up to 5:

DD DD D means 2 2 1
D DDD D means 1 3 1
DDDD D means 4 1
DDDDD means 5

So any grouping of the 5 letters DDDDD will give us a number the sum of whose digits is 5:

DD DD D will give us 221, 2021, 2201, or 2210.
etc.

So the question becomes:

How many ways can the 5 letters DDDDD be divided into groups?

We can have at most 4 groups, because if we divide DDDDD into 5 groups (D-D-D-D-D), we'll have a 5-digit number. Thus we can divide DDDDD at most 3 times. Let | = separation between groups.

D|D|D|DD = 1112
DD||D|DD = 2012
DD|D||DD = 2102
|DD|D|DD = 0212 = 212
DD|D|DD| = 2120
etc.

So any arrangement of the 8 elements DDDDD||| represents a number the sum of whose digits is 5.

Number of ways to arrange DDDDD||| = 8!/(5!*3!) = 56.

The correct answer is C.

By extension, the number of positive integers less than 10,000 the sum of whose digits is 4 = number of ways to arrange DDDD||| = 7!/(4!*3!) = 35.

Sorry, I don't get why you can't have D-D-D-D-D (1 1 1 1 1), that equals 5?