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What is the sum of all 3 digit numbers that leave a remainde
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Updated on: 26 Apr 2014, 08:51
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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
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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
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06 May 2014, 22:26
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
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06 Aug 2009, 19:52
The series will be of the form: 101, 104, 107.....995, 998.
It will have a total of 300 terms: 999100=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
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06 Aug 2009, 20:49
What is this AP formula? What does AP stand for? Thanks.



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Re: Sum & Remainder
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06 Aug 2009, 21:01
Arithmetic Progression!



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Re: Sum & Remainder
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07 Aug 2009, 01:38
Jivana wrote: The series will be of the form: 101, 104, 107.....995, 998.
It will have a total of 300 terms: 999100=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+(n1)d So here a = 101, d=3,n th term = 998 998 = 101 + (n1)3 ..on solving n = 300 ....so there r 300 termsin the series...



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Re: Sum & Remainder
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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 + (n1) 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
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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
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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
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11 May 2014, 04:14



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Re: What is the sum of all 3 digit numbers that leave a remainde
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28 May 2014, 05:51
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



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Re: What is the sum of all 3 digit numbers that leave a remainde
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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
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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
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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
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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
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27 Jul 2014, 13:53



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Re: What is the sum of all 3 digit numbers that leave a remainde
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22 Apr 2016, 16:33
Here is what i did => here the series is => 101+104+.....998 number of terms =300 (using An=A+(n1)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
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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
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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: 999100=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
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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 evenlyspaced set of integers, we can use the following equation to determine the sum: sum = average x quantity The smallest 3digit 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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