What is the sum of all remainders obtained when the first

Question

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

Bunuel · Accepted Answer

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.

Mindreko · Answer

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?

Macsen · Answer

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!!

shrikantv · Answer

Sum of remainer of natural nos. 1 to 100
(1+2+3+...8) * 11 +1
=8*9/2 * 11 + 1
= 397

mindmind · Answer

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....

mindmind · Answer

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...

Bunuel · Answer

Yes, that is correct.

noluyoyaa · Answer

(1+2+3+...+8)=36

This has 11 repetations in 100 plus 100 itself so 36*11+1=297

shreeya · Answer

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...

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 !!

drasticgre · Answer

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

FelixM · Answer

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!

Celestial09 · Answer

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

KarishmaB · Answer

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)

BrushMyQuant · Answer

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!

Watch the following video to MASTER Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

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What is the sum of all remainders obtained when the first

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