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What is the sum of all remainders obtained when the first

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Macsen
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What is the sum of all remainders obtained when the first [#permalink] New post   12 Nov 2011, 16:45
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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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A
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Bunuel
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Re: Sum of remainders when divided by 9 [#permalink] 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] New post   12 Nov 2011, 17:23
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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, 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] 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] 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] New post   19 Oct 2012, 09:41
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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, 09:47
Expert Reply
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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Re: What is the sum of all remainders obtained when the first [#permalink] 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] 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] 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] 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] 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] New post   03 Aug 2016, 23:08
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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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What is the sum of all remainders obtained when the first [#permalink] New post   27 Dec 2023, 03:02
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What is the sum of all remainders obtained when the first 100 natural numbers are divided by 9?

Remainders of the number from 1 to 100 when divided by 9 will
    ✤ Repeat in the pattern (1,2,3,4,5,6,7,8,0) and will
    ✤ Repeat \(\frac{100}{9}\) ~ 11 times for numbers from 1 to 99 and we will have an extra remainder of 1 which is from \(\frac{100}{9}\)


Remainder of 1 by 9 = 1
Remainder of 2 by 9 = 2
Remainder of 3 by 9 = 3
Remainder of 4 by 9 = 4
Remainder of 5 by 9 = 5
Remainder of 6 by 9 = 6
Remainder of 7 by 9 = 7
Remainder of 8 by 9 = 8
Remainder of 9 by 9 = 0

=> Sum of Remainders of numbers from 1 to 100 when divided by 9 = 11 * ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 0) + 1 = 11 * \(\frac{8*9}{2}\) + 1 = 11 * 36 + 1 = 397

So, Answer will be A
Hope it helps!

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