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705-805 (Hard)|   Remainders|      
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What is the sum of all 3 digit numbers that leave a remainde 25 Jul 2016, 00:54
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all three digit numbers 100~999

the first number that satisfies the description "leave a remainder of '2' when divided by 3" is 101, and the last one is 998.

total numbers that satisfy the description is:
n=[(998-101)/3]+1, and we have n=300

As the numbers are evenly spaced sequence, we could calculate the sum by:
S=n*mean
mean=(a1+an)/2, so mean=(101+998)/2=549.5

Finally S=300*549.5=146850

Thanks,
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What is the sum of all 3 digit numbers that leave a remainde 27 Oct 2016, 04:57
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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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What is the sum of all 3 digit numbers that leave a remainde 08 Nov 2016, 02:24
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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.
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Hi,
how did you get count as 300? as there is no common difference value is provided
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What is the sum of all 3 digit numbers that leave a remainde 09 Nov 2016, 11:01
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joyseychow
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

A. 897
B. 164,850
C. 164,749
D. 149,700
E. 156,720
Show more

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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What is the sum of all 3 digit numbers that leave a remainde 08 Apr 2018, 06:55
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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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What is the sum of all 3 digit numbers that leave a remainde 09 Apr 2018, 09:13
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From 1 to 99 we have 33 numbers which leave a remainder of 2 when divided by 3 (You call it as 1/3rd of the numbers) .
33 numbers leave a remainder of 0.
33 numbers leave a remainder of 1.
33 numbers leave a remainder of 2.
If we take three digit number, then try to find sum of 100 to 999 and divide by 3 (1/3rd).

(900/2)*(100+999) = 164850
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What is the sum of all 3 digit numbers that leave a remainde 27 Jan 2020, 19:05
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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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What is the sum of all 3 digit numbers that leave a remainde 27 Dec 2023, 05:06
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What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

First number = 101 (As 101 = 33*3 + 2)
Last Number = 998 (As 999 is divisible by 3)
Common difference, d = 3
Number of terms, n = (Last term - First Term)/d + 1 = \(\frac{998 - 101}{3}\) + 1 = 299 + 1 = 300
Sum = n * (First Term + Last Term)/2 = 300 * \(\frac{(101 + 998)}{2}\) = 164,850

So, Answer will be B
Hope it helps!

Watch the following video to MASTER Sequence problems

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What is the sum of all 3 digit numbers that leave a remainde 07 Jul 2025, 01:50
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