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

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What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post Updated on: 26 Apr 2014, 08:51
3
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A
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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

Originally posted by joyseychow on 06 Aug 2009, 19:33.
Last edited by Bunuel on 26 Apr 2014, 08:51, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 06 May 2014, 22:26
5
5
kanusha wrote:
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:)


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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Re: Sum & Remainder  [#permalink]

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New post 06 Aug 2009, 19:52
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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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Re: Sum & Remainder  [#permalink]

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New post 06 Aug 2009, 20:49
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What is this AP formula? What does AP stand for? Thanks.
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Re: Sum & Remainder  [#permalink]

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New post 06 Aug 2009, 21:01
Arithmetic Progression!
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Re: Sum & Remainder  [#permalink]

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New post 07 Aug 2009, 01:38
2
Jivana wrote:
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.


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...
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Re: Sum & Remainder  [#permalink]

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New post 07 Aug 2009, 12:31
bipolarbear wrote:
What is this AP formula? What does AP stand for? Thanks.


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)
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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 06 May 2014, 21:32
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:)
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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 10 May 2014, 09:37
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...
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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 11 May 2014, 04:14
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bytatia wrote:
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: \(1=9*0+1\).

Theory on remainders problems: remainders-144665.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199

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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 28 May 2014, 05:51
3
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
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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 04 Jun 2014, 00:24
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 :)
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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 11 Jun 2014, 04:10
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
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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 02 Jul 2014, 18:03
Valrus wrote:
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 ?
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What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 27 Jul 2014, 13:38
Narenn wrote:
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


Hi,

could you explain the logic in this formula?
my mistake was to take (last term - first term+1)/3
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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 27 Jul 2014, 13:53
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oss198 wrote:
Narenn wrote:
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


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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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 22 Apr 2016, 16:33
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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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 27 Oct 2016, 03:57
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  [#permalink]

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New post 08 Nov 2016, 01:24
Jivana wrote:
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.


Hi,
how did you get count as 300? as there is no common difference value is provided
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Re: What is the sum of all 3 digit numbers that leave a remainde  [#permalink]

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New post 09 Nov 2016, 10:01
joyseychow wrote:
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


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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