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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] New post 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
[Reveal] Spoiler: OA
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Re: Sum of remainders when divided by 9 [#permalink] New post 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] New post 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] New post 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] New post 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] New post 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.

Answer: A.

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

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 [#permalink] New post 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...


Yes, that is correct.
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Resources:
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: What is the sum of all remainders obtained when the first [#permalink] New post 09 Mar 2014, 11:37
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What is the sum of all remainders obtained when the first 100 natural? [#permalink] New post 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


Please post your answers with Explanation!
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Re: What is the sum of all remainders obtained when the first [#permalink] New post 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


Please post your answers with Explanation!


Search for a question before posting.

Locate the correct forum for your questions, choose either DS or PS forum. Do not post in general quantitative forum.

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


Topics merged.

Refer to solution above.
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Re: What is the sum of all remainders obtained when the first [#permalink] New post 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] New post 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] New post 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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What is the sum of all remainders obtained when the first   [#permalink] 01 Nov 2015, 23:43
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