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

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

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12 Nov 2011, 15:45
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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 [#permalink]

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12 Nov 2011, 16:23
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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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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
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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....
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Re: Sum of remainders when divided by 9 [#permalink]

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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:

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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19 Oct 2012, 08:41
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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...
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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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Re: What is the sum of all remainders obtained when the first [#permalink]

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09 Mar 2014, 11:37
Bumping for review and further discussion.
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Re: What is the sum of all remainders obtained when the first [#permalink]

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17 Jul 2015, 02:21
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What is the sum of all remainders obtained when the first 100 natural? [#permalink]

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12 Sep 2015, 09:57
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: What is the sum of all remainders obtained when the first [#permalink]

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12 Sep 2015, 13:19
hiteshahire22 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

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Follow ALL the posting guidelines (link in my signatures), including providing OA, 3 tags (source, difficulty and area).

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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
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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 !!
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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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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 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
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 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.

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] 03 Aug 2016, 22:08
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