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Please help! Remainder problem

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by dimkras » Sat Sep 04, 2010 12:11 am
Would appreciate your help with this one:

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

Tried to solve many times in many ways but I never get any of these answers.
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by sanju09 » Sat Sep 04, 2010 12:52 am
dimkras wrote:Would appreciate your help with this one:

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

Tried to solve many times in many ways but I never get any of these answers.

Remainders would appear in the repeating 11 cycles of 1, 2, 3, 4, 5, 6, 7, 8, and 0 uptil 99, and 1 will be the final remainder when 9 would divide 100.

Total of remainders = 11 × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 0) + 1 = 11 × 36 + 1 = [spoiler]397.


A

Choices are not in order, what's the source?[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
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by dimkras » Sat Sep 04, 2010 1:29 am
Thanks a lot, but can you give a little bit more detailed explanation of your logic. I don't get why there are 11 cycles.

Source is https://free-quiz.4gmat.com/quizzes/math_quiz_17/
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by sanju09 » Sat Sep 04, 2010 1:31 am
dimkras wrote:Thanks a lot, but can you give a little bit more detailed explanation of your logic. I don't get why there are 11 cycles.

Source is https://free-quiz.4gmat.com/quizzes/math_quiz_17/
9 could go 11 times in 100
The mind is everything. What you think you become. -Lord Buddha



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by dimkras » Sat Sep 04, 2010 1:36 am
Ah.. Now I see. Thanks a lot!
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