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# What is the sum of all remainders obtained when the first

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Intern
Joined: 15 Mar 2010
Posts: 11
What is the sum of all remainders obtained when the first  [#permalink]

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12 Nov 2011, 15:45
1
16
00:00

Difficulty:

35% (medium)

Question Stats:

75% (02:16) correct 25% (02:28) wrong based on 302 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 52906
Re: Sum of remainders when divided by 9  [#permalink]

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19 Oct 2012, 06:29
2
6
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.

Hope it's clear.
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Re: Sum of remainders when divided by 9  [#permalink]

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12 Nov 2011, 16:23
5
3
Remainers appear in the re-occuring 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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##### General Discussion
Intern
Joined: 15 Mar 2010
Posts: 11
Re: Sum of remainders when divided by 9  [#permalink]

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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!!
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Re: Sum of remainders when divided by 9  [#permalink]

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12 Nov 2011, 17:04
Sum of remainer of natural nos. 1 to 100
(1+2+3+...8) * 11 +1
=8*9/2 * 11 + 1
= 397
Intern
Joined: 23 May 2012
Posts: 29
Re: Sum of remainders when divided by 9  [#permalink]

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19 Oct 2012, 03:07
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....
Intern
Joined: 23 May 2012
Posts: 29
Re: Sum of remainders when divided by 9  [#permalink]

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19 Oct 2012, 08:41
1
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.

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...
Math Expert
Joined: 02 Sep 2009
Posts: 52906
Re: Sum of remainders when divided by 9  [#permalink]

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19 Oct 2012, 08:47
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.

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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Joined: 08 Sep 2015
Posts: 2
Re: What is the sum of all remainders obtained when the first  [#permalink]

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13 Sep 2015, 00:27
(1+2+3+...+8)=36

This has 11 repetations in 100 plus 100 itself so 36*11+1=297
Intern
Joined: 20 Apr 2015
Posts: 4
What is the sum of all remainders obtained when the first  [#permalink]

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11 Oct 2015, 08:52
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 !!
Intern
Joined: 08 Mar 2015
Posts: 2
What is the sum of all remainders obtained when the first  [#permalink]

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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
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Intern
Joined: 03 Feb 2016
Posts: 10
Re: What is the sum of all remainders obtained when the first  [#permalink]

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11 Feb 2016, 17:27
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.

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  [#permalink]

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03 Aug 2016, 19:00
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) a-b

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Re: What is the sum of all remainders obtained when the first  [#permalink]

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03 Aug 2016, 22:08
1
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) a-b

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

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Re: What is the sum of all remainders obtained when the first   [#permalink] 06 Feb 2019, 21:21
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# What is the sum of all remainders obtained when the first

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