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

Hope that's clear.
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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 can't recall, so I'm doing it in a crude way.}

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

= (101 + 998) / 2 * 300
= 1099 * 300 / 2
= 164, 840
So Ans = B

For these kind of problems, if one knows the AP formulas, then all needs to be done is setup a equation, and boom boom...
I do not think picking numbers will cut it for these types of PS.

What is this AP formula? What does AP stand for? Thanks.

Arithmetic Progression!

Formula for n th term = a+(n-1)d
So here a = 101, d=3,n th term = 998

998 = 101 + (n-1)3 ..on solving n = 300 ....so there r 300 termsin the series...

While Ap stands for Arithmetic Progression,

Nth term in the series = first term + (n-1) Common difference

Sum of the series = n/2 (First term + last term)

My Thought : Integer Rules
My Understanding of this Question
Find the number , Upon Sum of 3 Digits of a number Gives a Reminder 2 when it is Divided by 3
Seeing the Options After Dividing an Finding the Reminder of 2
My Answer was C
My Answer was Wrong

But I Don't Understand Why there is need of AP , What's the Question is Testing

Pls Help:)

Hello. I'm sorry I'm just not sure in which section to post this kind of question.

I just have a question about remainders. When 2 is divided by 7, how come the remainder is 2? As 2/7 is 0.285714...

Let me ask you a question: how many leftover apples would you have if you had 2 apples and wanted to distribute in 7 baskets evenly? Each basket would get 0 apples and 2 apples would be leftover (remainder).

When a divisor is more than dividend, then the remainder equals to the dividend, for example:
3 divided by 4 yields the reminder of 3: \(3=4*0+3\);
9 divided by 14 yields the reminder of 9: \(9=14*0+9\);
1 divided by 9 yields the reminder of 1: \(1=9*0+1\).


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Should be straightforward...

101....104...107...110....998

Now, first lets find the number of terms

998 - 101 = 897 / 3 = 299 + 1 = 300 terms

Now then, lets find the average

998 + 101 = 549.5

Now multiply (549.65)(300)= 164,850

Answer is B

Hope this clarifies
Cheers
J

Reading through question is key here :
Given sum of three digit numbers : the smallest three digit is 100 but the condition is when number is divided by 3 the remainder must be 2 hence the series will begin from 101 till the largest three digit number (when divided by 3 gives remainder 2) i.e. 998.

to find the sum of AP : N/2 (a + l) where N is number of terms , a is first term here its 101 and l is last term of series here its 998.
to calculate N = (last term - first number)/3 + 1
hence N = 300
Sum = 300/2 (101 + 998)
Hence answer is B

Queries are encouraged

The right answer could be found by 5 steps:
1. The First term - 101 and the last one - 998
2. Their sum - 1099 and their difference - 897
3. The number of terms: 897/3+1 = 300
4. The number of pairs: 300/2 = 150
5. Sum of pairs = 150*1099 = 164850

Hence B

In step 3, why are we dividing by 3 + 1 ?

Hi,

could you explain the logic in this formula?
my mistake was to take (last term - first term+1)/3

The following post might help: how-many-multiples-of-4-are-there-between-12-and-94862.html#p730075
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First term in AP,a = 101 = 33*3 + 2
Last term in AP,l = 998= 332*3 + 2
Number of terms in sequence
= 332- 33+1
= 300
Sum of n terms in AP
= number of terms/2 *[a+l]
= 300/2 *[101 + 998]
=150 * 1099
=164850
Answer B
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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.