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
Counting
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- talaangoshtari
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NOTE: I have replaced talaangoshtari's answer choices with the answer choices that appeared in the original question.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
The answer is C
More on this question here: https://www.beatthegmat.com/very-tricky- ... 25349.html
Cheers,
Brent
- MartyMurray
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Positive numbers less than 10,000 can have one to four digits.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
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
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Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.
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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
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
- talaangoshtari
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