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Expert replies
by talaangoshtari » Sat May 16, 2015 8:44 am
How many positive integers less than 10,000 are such that the sum of their digits is 5?

A.56
B.22
C.100
D.75
E.80
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Source: — Problem Solving |

by Brent@GMATPrepNow » Sat May 16, 2015 8:49 am
talaangoshtari wrote:How many positive integers less than 10,000 are such that the sum of their digits is 5?

(a) 31
(b) 51
(c) 56
(d) 62
(e) 93
NOTE: I have replaced talaangoshtari's answer choices with the answer choices that appeared in the original question.

The answer is C

Image


More on this question here: https://www.beatthegmat.com/very-tricky- ... 25349.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by MartyMurray » Fri May 22, 2015 9:30 pm
talaangoshtari wrote:How many positive integers less than 10,000 are such that the sum of their digits is 5?

(a) 31
(b) 51
(c) 56
(d) 62
(e) 93
Positive numbers less than 10,000 can have one to four digits.

One digit numbers whose digits add up to 5 can only use the number 5.

5

So there is just 1 one digit number whose digits add up to 5.

Two digit numbers whose digits add up to 5 can be permutations of (1,4), (2,3), and (5,0).

Permutations of two elements have two arrangements, except for (5,0) which can only have one, 50, as 05 is not really a two digit number.

So 2 + 2 + 1 = 5 two digit numbers.

Three digit numbers whose digits add up to five can be permutations of (1,1,3), (0,1,4), (0,2,3), (0,0,5), and (2,2,1).

For example (1,1,3) can be arranged 3!/2! = 3 ways.

However, because of the zeros, (0,0,5) can only be arranged one way, as 500.

Taking into account the doubles and the zeros, one can arrange those sets in the following number of ways respectively.

3 + 4 + 4 + 1 + 3 = 15 three digit numbers.

Four digit numbers whose digits add up to five can be permutations of (1,1,3,0), (0,0,1,4), (0,0,2,3), (0,0,0,5), (0,2,2,1) and (1,1,1,2).

For example, (1,1,3,0) can be arranged (3 x 3 x 2 x 1)/2! = 9 ways.

(1,1,1,2) can be arranged 4!/3! = 4 ways.

Taking into the account the doubles, triples and zeros, one can arrange those sets in the following number of ways respectively.

9 + 6 + 6 + 1 + 9 + 4 = 35

So the total of one digit, two digit, three digit, and four digit numbers is the following.

1 + 5 + 15 + 35 = 56

Choose C.
Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
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by nikhilgmat31 » Mon May 25, 2015 1:22 am
Very Good Question, What is the source of such questions.
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by Brent@GMATPrepNow » Mon May 25, 2015 7:04 am
I "created" this question in 2008 (which was based on another resource I had created earlier): https://www.beatthegmat.com/very-tricky- ... 25349.html

I had seen a similar question in one of my combinatorics classes in university.

This question is skirting the acceptable limits of the GMAT. Probably 800+

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by talaangoshtari » Mon May 25, 2015 6:19 pm
I found this problem in my high school book.
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by nikhilgmat31 » Mon May 25, 2015 7:53 pm
Good question, keep posting such questions
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