What is the sum of all 3 digit numbers that leave a remainde

The series is => 101+...998
Number of terms => 300
sum => 300/2 * [101+998]
hence sum => 164850 i.e. option B
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Here is what i did =>
here the series is => 101+104+.....998 number of terms =300 (using An=A+(n-1)D i.e. nth term of an AP formula)
Sum = 300/2 * (101+998)= 150*1099
Now Hold ON do not multiply
The last digit here is 0 and second last is 5 => Smash that B
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Hi Bunnel,

I solved this question using the equations of Arithmetic progression. Is there an alternate way so that long multiplication can be avoided?

Thanks.

Another approach to this question =>
clearly the question asks us to find the sum of series => 101+104+107....998
here the number of terms => [(first term - last term)/3]+ 1 = 300
Mean of any AP series is the average of the first term and the last term
hence here the mean => (101+998)/2 = 1099/2
now mean = sum of all the terms / total no. of terms
hence the sum of all the terms = mean * total number of terms.
hence the sum = (1099/2) *300 = 164850
hence B
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Hi,
how did you get count as 300? as there is no common difference value is provided

As we can see, the integers in question are: 101, 104, 107, …, 998. Since we are calculating a sum of an evenly-spaced set of integers, we can use the following equation to determine the sum:

sum = average x quantity

The smallest 3-digit integer that leaves a remainder of 2 when divided by 3 is 101, and the largest is 998. We can use these values to determine the average and quantity. Let’s start by determining the average:

avg = (largest integer + smallest integer)/2

avg = ( 998 + 101)/2

avg = 1,099/2

Next we can determine the quantity:

quantity = (largest integer - smallest integer)/3 + 1

quantity = (998 - 101)/3 + 1 = 299 + 1 = 300

Now we can determine the sum:

sum = (1,099/2)(300) = 164,850

Answer: B
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The use of arithmetic formula will help you find the sum of such integers quickly.

We want all three digit integers that gives 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. You 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 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 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



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