How many 3-digit numbers have the sum of their digits equal to 4?

Question

A. 4
B. 6
C. 7
D. 10
E. 12

CrackverbalGMAT · Accepted Answer

For a 3-digit number to have the sum of its digits equal to 4 (exactly), it can only begin with 1, 2, 3 or 4.

Case 1: Number begins with 1

1  _ _: remaining digits can be:

1a: 0 & 3 --> 2 such numbers
1b: 1 & 2 --> 2 such numbers

Therefore total number of such numbers possible = 4

Case 2: Number begins with 2

2  _ _: remaining digits can be:

2a: 0 & 2 --> 2 such numbers
2b: 1 & 1 --> 1 such number

Therefore total number of such numbers possible = 3

Case 3: Number begins with 3

3  _ _: remaining digits can  only  be 0 & 1 --> 2 such numbers possible

Case 4: Number begins with 4

4  _ _: remaining digits can  only  be 0 & 0 --> 1 such number possible

Therefore total number of such numbers possible = 4 + 3 + 2 + 1 =  10

Answer: (D)

Hope this helps.

rajatchopra1994 · Answer

Number of cases:

103, 202, 301, 112, 121, 211, 130, 400, 310, 220.

Total Cases = 10

IMO-D

InaKi20 · Answer

Is there any faster method that can be applied?

TestPrepUnlimited · Answer

We can separate into possible digits to select first then scramble the digits.
For example, 4 + 0 + 0 is one case, only 400 suffices being a 3-digit number.
3 + 1 + 0 -> 310, 301, 130, 103.
2 + 2 + 0 -> 220, 202.
2 + 1 + 1 -> 211, 121, 112.

Those are the only cases, thus there are 10 possible numbers. Notice how we break down from large digits into small digits one step at a time.

Ans: D

ElninoEffect · Answer

Factorial might help.!

Digits summing to four = 211, 310 & 400.

211 can be rearranged in 3!/2! ways = 3.
310 can be rearranged in 3! ways = 6.
400 can be rearranged in  one way  only as replacing 4 from 1st position will lead to the failure of the condition i.e 'It should be a three digit integer'.

Total Ways :- 6+3+1 = 10.

MS61 · Answer

yes ! Watch this video for the quiekest way to solve this type of problem  (1) Permutations and Combinations 10 (Similar to Different Distribution) - YouTube

MihirBathia · Answer

Looking at the options I got the answer, but in such problems, I am always on the verge of getting on my horse and calculating the number of terms in the AP, where a(first term)=103, d(difference)=9 and tn (last term)=994. The problem is, I understand 994 also to be a number with the sum 4=9+9+4, 22=2+2=4. Basically I end up calculating the seed of the number. Please help

chetanraghav17 · Answer

It seems you made an error when saying 310 can be rearranged in 3! Ways, as the leftmost digit can’t be 0. Also, you missed one more case where digits could be 220.

Posted from my mobile device

Gemmie · Answer

4 = 4 + 0 + 0 = 3 + 1 + 0 = 2 + 2 + 0 = 2 + 1 + 1

4, 0, 0:  1 way
 3, 1, 0:  2 * 2 = 4 ways
 2, 2, 0:  2 ways (220 and 202)
 2, 1, 1:   \frac{3!}{2!} = 3  ways

==> total: 1 + 4 + 2 + 3 = 10

luisdicampo · Answer

Deconstructing the Question

We are looking for 3-digit numbers such that the sum of the digits is 4.

Let the digits be  a, b, c , where  a \ge 1  since it is a 3-digit number.

We solve:

a + b + c = 4

Step-by-step

Case  a = 1 :

b + c = 3

Solutions:  (0,3), (1,2), (2,1), (3,0)

Total:  4

Case  a = 2 :

b + c = 2

Solutions:  (0,2), (1,1), (2,0)

Total:  3

Case  a = 3 :

b + c = 1

Solutions:  (0,1), (1,0)

Total:  2

Case  a = 4 :

b + c = 0

Solution:  (0,0)

Total:  1

Total numbers:

4 + 3 + 2 + 1 = 10

Answer D

How many 3-digit numbers have the sum of their digits equal to 4?

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GMAT Problem Solving (PS) Questions

How many 3-digit numbers have the sum of their digits equal to 4?

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