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Re: Sum of remainders when divided by 9 [#permalink]
12 Nov 2011, 16:30
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!!
Re: Sum of remainders when divided by 9 [#permalink]
19 Oct 2012, 06:29
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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:
Re: Sum of remainders when divided by 9 [#permalink]
19 Oct 2012, 08:41
1
This post was BOOKMARKED
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...
Re: Sum of remainders when divided by 9 [#permalink]
19 Oct 2012, 08:47
Expert's post
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...
Re: What is the sum of all remainders obtained when the first [#permalink]
17 Jul 2015, 02:21
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What is the sum of all remainders obtained when the first [#permalink]
01 Nov 2015, 23:43
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 _________________
Happy Learning/ DrasticGRE .................
gmatclubot
What is the sum of all remainders obtained when the first
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01 Nov 2015, 23:43
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