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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 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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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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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 19 Oct 2012, 07:29
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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]

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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]

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New post 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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Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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What is the sum of all remainders obtained when the first 100 natural? [#permalink]

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New post 12 Sep 2015, 10:57
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]

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New post 12 Sep 2015, 14:19
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]

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 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) a-b

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

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New post 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) a-b

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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Re: What is the sum of all remainders obtained when the first   [#permalink] 03 Aug 2016, 23:08
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