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Basically, the question asks how many 4 digit numbers (including those in the form 0XXX, 00XX, and 000X) have digits which add up to 5. Think about the question this way: we know that there is a total of 5 to be spread among the 4 digits, we just have to determine the number of ways it can be spread.

Let X represent a sum of 1, and | represent a seperator between two digits. As a result, we will have 5 X's (digits add up to the 5), and 3 |'s (3 digit seperators).

So, for example:

XX|X|X|X = 2111 ||XXX|XX = 0032

etc.

There are 8C3 ways to determine where to place the separators. Hence, the answer is 8C3 = 56.

Basically, the question asks how many 4 digit numbers (including those in the form 0XXX, 00XX, and 000X) have digits which add up to 5. Think about the question this way: we know that there is a total of 5 to be spread among the 4 digits, we just have to determine the number of ways it can be spread.

Let X represent a sum of 1, and | represent a seperator between two digits. As a result, we will have 5 X's (digits add up to the 5), and 3 |'s (3 digit seperators).

So, for example:

XX|X|X|X = 2111 ||XXX|XX = 0032

etc.

There are 8C3 ways to determine where to place the separators. Hence, the answer is 8C3 = 56.

This is correct. Also this is the best way to solve this question. +1. (Solved exactly the same way)

Since the sum of the 4 digits have to equal to 5, anyting #s >=6000 is eliminated.

5000s range - 1 5000 is the only possibly number in the 5000's range.

4000s range - 3 Since 4+1=5, there are only 3 spots to put the 1s, namely 4100, 4010, 4001.

3000s range - 6 We can have 3+2+0+0 (3 combinations) or 3+1+1+0 (3 combinations).

2000s range - 7 We can have 2+2+1+0 (6 combinations because of 3! ways of arranging the last 3 digits using 2,1,0) or 2+1+1+1 (1 combination).

1000s range - 15 We can have 1+1+1+2 (3 combinations), 1+2+2+0 (3 combinations), 1+4+0+0 (3 combinations), and 1+3+1+0 (6 combinations because of 3! ways of arranging the last 3 digits of 3,1,0).

The # of numbers is therefore 1+3+6+7+15=32. Notice that you get 3 combinations whenever 2 of the last 3 digits are the same.

Then use the same logic on 3 digits and 2 digits and 1 digit numbers.

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

Consider this: we have 5 \(d\)'s and 3 separators \(|\), like: \(ddddd|||\). How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(d\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\).

With these permutations we'll get combinations like: \(|dd|d|dd\) this would be 3 digit number 212 OR \(|||ddddd\) this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR \(ddddd|||\) this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)...

Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5.

Hence the answer is \(\frac{8!}{5!3!}=56\).

Answer: C (56).

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56\)
_________________

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

Consider this: we have 5 \(d\)'s and 3 separators \(|\), like: \(ddddd|||\). How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(d\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\).

With these permutations we'll get combinations like: \(|dd|d|dd\) this would be 3 digit number 212 OR \(|||ddddd\) this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR \(ddddd|||\) this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)...

Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5.

Hence the answer is \(\frac{8!}{5!3!}=56\).

Answer: C (56).

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56\)

there is a shortcut. For the problem, 4 digits are equally important in 0000-9999 set and it is impossible to build a number using only one digit (like 11111) So, answer has to be divisible by 4. Only 56 works.

"There are 8C3 ways to determine where to place the separators"

I'm not familiar with this shortcut/approach.

Ta

Consider this: we have 5 \(d\)'s and 3 separators \(|\), like: \(ddddd|||\). How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(d\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\).

With these permutations we'll get combinations like: \(|dd|d|dd\) this would be 3 digit number 212 OR \(|||ddddd\) this would be single digit number 5 (smallest number less than 10,000 in which sum of digits equals 5) OR \(ddddd|||\) this would be 4 digit number 5,000 (largest number less than 10,000 in which sum of digits equals 5)...

Basically this arrangements will give us all numbers less than 10,000 in which sum of the digits (sum of 5 d's=5) equals 5.

Hence the answer is \(\frac{8!}{5!3!}=56\).

Answer: C (56).

This can be done with direct formula as well:

The total number of ways of dividing n identical items (5 d's in our case) among r persons or objects (4 digt places in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r-1}_C_{r-1}\).

In our case we'll get: \({n+r-1}_C_{r-1}={5+4-1}_C_{4-1}={8}C3=\frac{8!}{5!3!}=56\)

Also see the image I found in the net about this question explaining the concept:

Attachment:

pTNfS-2e270de4ca223ec2741fa10b386c7bfe.jpg [ 63.83 KiB | Viewed 62021 times ]

Bunnel I have solved it in this way but your methods seems to be quicker. But I couldn't understand. Could you please explain in simple words.
_________________

Bunnel I have solved it in this way but your methods seems to be quicker. But I couldn't understand. Could you please explain in simple words.

this approach was more natural and my first idea to solve the promblem

Then I saw the Bunnel´s and AKProdigy87´s way to solve this problem.

The idea of Bunnel and AKProdigy87 method is that the sum of the digits must equal 5, and this five can be distributed among the 4 digits and numbers are made by "ones". Again, numbers are made by "ones"! please do not thing of numbers as 2, 3, 4 or 5 digit for example:

Can this method be used in variations of this question?

Such as sum of 7 digits that add to equal 6?

Yes. Why not? You mean numbers with 7 digits? I don't know if that is going to be a question of gmat, but... If the question was to equal 5 you can do the same to six, just add one X