What is the Sum of all 3 digit Numbers that Leave a Remainder of '2' when Divided by 3? GMAT Problem Solving

Question: What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

1. 897
2. 164,850
3. 164,749
4. 149,700
5. 156,720

Correct Answer: B

Solution and Explanation
Approach Solution 1:

The use of arithmetic formulas will help you find the sum of such integers quickly.

We want all three digit integers that give the remainder of 2 upon dividing by 3.

some important points regarding this problem.

1) there are only 3 possible remainders for any integer upon dividing by 3 which are 0, 1, 2. Candidates will get the remainder of 2 when the integer being divided is 1 less than the integer that is evenly divisible by 3. First of such kind of integer in 101 and last is 998 (since 102 and 999 are evenly divisible by 3)

2) the next integer that gives the remainder of 2 when divided by 3 after 101 will be 101+3 i.e. 104. We can form here the sequence of all the integers that gives the remainder of 2 when divided by 3 in which every integer will be 3 greater than the previous integer. 101, 104, 107, 110, ...........,998.

3) So we have arithmetic progression with first term 101, last term 998, and common difference 3.

4) Formula to calculate number of terms in the sequence with common difference 3 is
(i) Including both ends [(last term - first term)/3] + 1
(ii) Including only one end [(last term - first term)/3]
(iii) Excluding both ends [(last term - first term)/3] - 1

5) Since we want both ends to be included in the counting, we will use the first case. So number of terms is [(998 - 101)/3] +1 ---------> 299 + 1 -------> 300

6) Sum of arithmetic progression with first term a, last term l and number of terms n will be n(a+l)/2 -------> 300(1099)/2 ---------> 164850

Hence, the correct answer is B.

Approach Solution 2

The series will be of the form: 101, 104, 107.....995, 998.

It will have a total of 300 terms: 999-100=899 + (1). = 900 (Take 1/3 of this, since only 1 term is there in every 3.) {There is a proper AP formula for this, but I can't recall, so it is done in a crude way.}

Now, sum = (1st number + nth number)/2 * n

= (101 + 998) / 2 * 300
= 1099 * 300 / 2
= 164, 850

So Answer is option B

For these kinds of problems, if one knows the AP formulas, then all that needs to be done is to set up an equation.

Approach Solution 3:
The lowest 3 digit number is 101 and hightest number is 998. 

Therefore, the AP is 101,104,107,….998 where, the first term is a= 101, the common difference is d= 104-101= 3 and the nth term is an= 998.

Now, let us find the number of terms

We know that nth term of AP(Arithmetic Progression)is given by 

an= a+(n-1)d

Now, substituting the values, we get 

an= a+(n-1)d
=> 998= 101+(n-1)3
=> (n-1)3= 998-101
=> (n-1)3= 897
=> n-1= 897/3
=> n-1= 299
=> n= 299+1
=> n= 300

W also know that the sum to nth term in an AP is given by,

S= n/2(1st+lstr term)

Therefore, we have

S= 300/2(101+998)= 150*1099= 164850 
 

“What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?”- is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

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