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What is the sum of all remainders obtained when the first
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12 Nov 2011, 16:45
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What is the sum of all remainders obtained when the first 100 natural numbers are divided by 9? A. 397 B. 401 C. 403 D. 405 E. 399
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Re: Sum of remainders when divided by 9
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19 Oct 2012, 07:29
Macsen wrote: What is the sum of all remainders obtained when the first 100 natural numbers are divided by 9?
A. 397 B. 401 C. 403 D. 405 E. 399 A positive integer can give only the following 9 remainders when divided by 9: 1, 2, 3, 4, 5, 6, 7, 8, and 0. 1 divided by 9 gives the remainder of 1; 2 divided by 9 gives the remainder of 2; ... 8 divided by 9 gives the remainder of 8; 9 divided by 9 gives the remainder of 0. We'll have 11 such blocks, since 99/9=11. The last will be: 91 divided by 9 gives the remainder of 1; 92 divided by 9 gives the remainder of 2; ... 98 divided by 9 gives the remainder of 8; 99 divided by 9 gives the remainder of 0. The last number, 100, gives the remainder of 1 when divided by 9, thus the sum of all remainders will be: 11(1+2+3+4+5+6+7+8+0)+1=397. Answer: A. Hope it's clear.
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Re: Sum of remainders when divided by 9
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12 Nov 2011, 17:23
Remainers appear in the reoccuring cycles of 1,2,3,4,5,6,7,8 until 99, and then 1 for the 100.
11 * (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) + 1 = 397
Does that help?




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Re: Sum of remainders when divided by 9
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12 Nov 2011, 17:30
Thanks Mindreko. Got it!! However, I was starting off from 11 and hitting 8 such sets of natural numbers till 100 and the sum of which was 36 for each set. I was wrong. Thanks!!



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Re: Sum of remainders when divided by 9
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12 Nov 2011, 18:04
Sum of remainer of natural nos. 1 to 100 (1+2+3+...8) * 11 +1 =8*9/2 * 11 + 1 = 397



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Re: Sum of remainders when divided by 9
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19 Oct 2012, 04:07
This sum is interesting..
But even after reading the solutions .. I am not getting the concept right..
Can any one please explain this to me....



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Re: Sum of remainders when divided by 9
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19 Oct 2012, 09:41
Bunuel wrote: Macsen wrote: What is the sum of all remainders obtained when the first 100 natural numbers are divided by 9?
A. 397 B. 401 C. 403 D. 405 E. 399 A positive integer can give only the following 9 remainders when divided by 9: 1, 2, 3, 4, 5, 6, 7, 8, and 0. 1 divided by 9 gives the remainder of 1; 2 divided by 9 gives the remainder of 2; ... 8 divided by 9 gives the remainder of 8; 9 divided by 9 gives the remainder of 0. We'll have 11 such blocks, since 99/9=11. The last will be: 91 divided by 9 gives the remainder of 1; 92 divided by 9 gives the remainder of 2; ... 98 divided by 9 gives the remainder of 8; 99 divided by 9 gives the remainder of 0. The last number, 100, gives the remainder of 1 when divided by 9, thus the sum of all remainders will be: 11(1+2+3+4+5+6+7+8+0)+1=397. Answer: A. Hope it's clear. Thanks Bunuel!! To make my concepts more clear... Eg : What is the sum of the remainder, when the first natural nos are divided by 7... So it would be = (1+2+3+4+5+6)*14 +1+2 = 297 I hope this is right...



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Re: Sum of remainders when divided by 9
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19 Oct 2012, 09:47
mindmind wrote: Bunuel wrote: Macsen wrote: What is the sum of all remainders obtained when the first 100 natural numbers are divided by 9?
A. 397 B. 401 C. 403 D. 405 E. 399 A positive integer can give only the following 9 remainders when divided by 9: 1, 2, 3, 4, 5, 6, 7, 8, and 0. 1 divided by 9 gives the remainder of 1; 2 divided by 9 gives the remainder of 2; ... 8 divided by 9 gives the remainder of 8; 9 divided by 9 gives the remainder of 0. We'll have 11 such blocks, since 99/9=11. The last will be: 91 divided by 9 gives the remainder of 1; 92 divided by 9 gives the remainder of 2; ... 98 divided by 9 gives the remainder of 8; 99 divided by 9 gives the remainder of 0. The last number, 100, gives the remainder of 1 when divided by 9, thus the sum of all remainders will be: 11(1+2+3+4+5+6+7+8+0)+1=397. Answer: A. Hope it's clear. Thanks Bunuel!! To make my concepts more clear... Eg : What is the sum of the remainder, when the first 100 natural nos are divided by 7... So it would be = (1+2+3+4+5+6)*14 +1+2 = 297 I hope this is right... Yes, that is correct.
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Re: What is the sum of all remainders obtained when the first
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13 Sep 2015, 01:27
(1+2+3+...+8)=36
This has 11 repetations in 100 plus 100 itself so 36*11+1=297



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What is the sum of all remainders obtained when the first
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11 Oct 2015, 09:52
To make my concepts more clear...
Eg : What is the sum of the remainder, when the first 100 natural nos are divided by 7...
So it would be = (1+2+3+4+5+6)*14 +1+2 = 297 I hope this is right...[/quote]
hii, Can u please explain this concept,
Eg : What is the sum of the remainder, when the first 100 natural nos are divided by 7...
So it would be = (1+2+3+4+5+6)*14 +1+2
Thanks !!



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What is the sum of all remainders obtained when the first
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02 Nov 2015, 00:43
every consecutive 9 integer from 1, which divided by 9 gives a total remainder of 36... 1+2+...8+0=36 this will repeat 11 times till 99, we have 100 also which give another 1 as remainder. so, (36*11) +1= 397 Ans. A
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Re: What is the sum of all remainders obtained when the first
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11 Feb 2016, 18:27
mindmind wrote: Bunuel wrote: Macsen wrote: What is the sum of all remainders obtained when the first 100 natural numbers are divided by 9?
A. 397 B. 401 C. 403 D. 405 E. 399 A positive integer can give only the following 9 remainders when divided by 9: 1, 2, 3, 4, 5, 6, 7, 8, and 0. 1 divided by 9 gives the remainder of 1; 2 divided by 9 gives the remainder of 2; ... 8 divided by 9 gives the remainder of 8; 9 divided by 9 gives the remainder of 0. We'll have 11 such blocks, since 99/9=11. The last will be: 91 divided by 9 gives the remainder of 1; 92 divided by 9 gives the remainder of 2; ... 98 divided by 9 gives the remainder of 8; 99 divided by 9 gives the remainder of 0. The last number, 100, gives the remainder of 1 when divided by 9, thus the sum of all remainders will be: 11(1+2+3+4+5+6+7+8+0)+1=397. Answer: A. Hope it's clear. Thanks Bunuel!! To make my concepts more clear... Eg : What is the sum of the remainder, when the first natural nos are divided by 7... So it would be = (1+2+3+4+5+6)*14 +1+2 = 297 I hope this is right... Hi, For my clarification as well, regarding the case in which we divide by 7, the first block is from 1 to 7, before the cycle starts again. Therefore, in order to find out how many full cycles of (from 0 to 6 as remainder) are we divide 100 by 7 (the number of possible remainders), which gives us 14 and a remainder of 2, which are the 1 and 2 remainders. Please let me know. Thank you!



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Re: What is the sum of all remainders obtained when the first
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03 Aug 2016, 20:00
Hi! I am no math geek.. so i thought in simple terms.. but couldn't arrive at the answer ... please help me in analysis my approach step 1) sum of 1st 100 natural numbers = a step 2) add 9+18=+27+36+45+54+63+72+81+90+99 =b step 3) ab
please help



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Re: What is the sum of all remainders obtained when the first
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03 Aug 2016, 23:08
Celestial09 wrote: Hi! I am no math geek.. so i thought in simple terms.. but couldn't arrive at the answer ... please help me in analysis my approach step 1) sum of 1st 100 natural numbers = a step 2) add 9+18=+27+36+45+54+63+72+81+90+99 =b step 3) ab
please help You need the sum of remainders. You don't have to add the 100 numbers. You have to divide each number by 9 and get the remainder. Then you have to add those 100 remainders. The question asks for the sum of the 100 remainders. So think about it. 1/9 remainder 1 2/9 remainder 2 and so on... till you get remainders 1, 2, 3,... 8, 0 for first 9 numbers. For next 9 numbers, again, remainders will be 1, 2, 3, ... 8, 0. This will happen 11 times till you reach the number 99. 100 will give you a remainder of 1. 1 + 2 + 3 + ... 8 = 8*9/2 = 36 Total sum = 11*36 + 1 = 397 Answer (A)
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