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How many positive integers less than 1000 are there in which the sum of the digits equals 4?
(A) 10
(B) 12
(C) 13
(D) 14
(E) 15
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
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(A) 10
(B) 12
(C) 13
(D) 14
(E) 15
Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
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Posts: 1,382
1 digit number - 4
count 1
2 digit number - 40,31,13,22
count 4
3 digit number in 100: 1 _ _
Possible combinations for tens and units digit: 12,21,03,30
count 4
3 digit number in 200: 2 _ _
Possible combinations for tens and units digit: 02,20,11
count 3
3 digit number in 300: 3 _ _
Possible combinations for tens and units digit: 01,10
count 2
3 digit number in 400: 4 _ _
Possible combinations for tens and units digit: 00
count 1
Total = 1+4+4+3+2+1 = 15
count 1
2 digit number - 40,31,13,22
count 4
3 digit number in 100: 1 _ _
Possible combinations for tens and units digit: 12,21,03,30
count 4
3 digit number in 200: 2 _ _
Possible combinations for tens and units digit: 02,20,11
count 3
3 digit number in 300: 3 _ _
Possible combinations for tens and units digit: 01,10
count 2
3 digit number in 400: 4 _ _
Possible combinations for tens and units digit: 00
count 1
Total = 1+4+4+3+2+1 = 15
Bunuel
shishimaru
Joined: 19 Aug 2024
Last visit: 22 Feb 2026
Posts: 159
Location: India
Concentration: Strategy, Technology
GPA: 2.2
WE:Sales (Finance)
Posts: 159
1-digit (1 to 9): Only 4 → 1 number.
2-digit (10 to 99): First digit 1 to 9, second 0 to 9, sum = 4.
- First = 1, Second = 3 → 13.
- First = 2, Second = 2 → 22.
- First = 3, Second = 1 → 31.
- First = 4, Second = 0 → 40. → 4 numbers.
3-digit (100 to 999): First 1 to 9, second & third 0 to 9, sum = 4.
- First = 1: Second + Third = 3 → 103, 112, 121, 130 → 4.
- First = 2: Second + Third = 2 → 202, 211, 220 → 3.
- First = 3: Second + Third = 1 → 301, 310 → 2.
- First = 4: Second + Third = 0 → 400 → 1. → Total 10.
All together: \(1 + 4 + 10 = 15\).
Answer: E
2-digit (10 to 99): First digit 1 to 9, second 0 to 9, sum = 4.
- First = 1, Second = 3 → 13.
- First = 2, Second = 2 → 22.
- First = 3, Second = 1 → 31.
- First = 4, Second = 0 → 40. → 4 numbers.
3-digit (100 to 999): First 1 to 9, second & third 0 to 9, sum = 4.
- First = 1: Second + Third = 3 → 103, 112, 121, 130 → 4.
- First = 2: Second + Third = 2 → 202, 211, 220 → 3.
- First = 3: Second + Third = 1 → 301, 310 → 2.
- First = 4: Second + Third = 0 → 400 → 1. → Total 10.
All together: \(1 + 4 + 10 = 15\).
Answer: E
Posts: 198
1 to 9= 1 digit 4
10 to 99= 40, 13,31,22= 4
100-199= 103, 121, 112, 130= 4
200-299= 202,220, 211= 3
300-399= 301,310=2
400-999= 400= 1
total= 1+4+4+3+2+1 =15
Correct answer is E
10 to 99= 40, 13,31,22= 4
100-199= 103, 121, 112, 130= 4
200-299= 202,220, 211= 3
300-399= 301,310=2
400-999= 400= 1
total= 1+4+4+3+2+1 =15
Correct answer is E
jahnvi01
Joined: 20 May 2025
Last visit: 15 Apr 2026
Posts: 1
Location: India
Schools: ISB '26 (A$$)
GMAT Focus 1: 715 Q90 V83 DI84 
GPA: 3.9
Number should be less than 1000 that means it has less than or equal to 3 digits.
We can use the Separator method here.
For number abc,
a+b+c = 4
There should be 2 separators => (4+2)!/4!*2! = 15
We can use the Separator method here.
For number abc,
a+b+c = 4
There should be 2 separators => (4+2)!/4!*2! = 15
conditions: 3 spaces and sum=4
so possibilities: 0+0+4, 0+1+3, 0+2+2, 1+1+2
0,0,4------3!/2!
0,1,3------3!
0,2,2------3!/2!
1,1,2------3!/2!
---------sum=15
so possibilities: 0+0+4, 0+1+3, 0+2+2, 1+1+2
0,0,4------3!/2!
0,1,3------3!
0,2,2------3!/2!
1,1,2------3!/2!
---------sum=15
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