The answer is 136.

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I had added 1234 into 2013 = 3247 then divided 3247 by 183 and got 136

Pirah Sikandar
- 7 years, 9 months ago

Its a really simple problem in reality but the thing is...u have 2 no how to solve it. :D

Siddharth Balakrishnan
- 7 years, 9 months ago

sir great logic

Ramesh Khinchi
- 7 years, 9 months ago

Take N=1234. N%183 = 136. Therefore, 136 is the answer.

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Easier. (2013*n+1234)/183 = 136

Philip John
- 7 years, 9 months ago

This seems to be true for all N that are congruent to 1234 mod 2013. Why is that the case?

Connor Stack
- 7 years, 9 months ago

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You should be careful that others can understand what you are writing. Definitions that you use should be clear, and notation should conform with that given in the problem (where possible).

I would disagree with the claim that "N is exactly divisible by 2013", as this contradicts the first statement of the problem. Are you thinking of some other value?

I apologize for the blunder made... It will be (N-1234) which is exactly divisible by 2013. Moreover as 183 is also a factor of 2013 it will completely divide (N-1234) leaving no remainder. But 183 being smaller than 2013 can divide the term, N, further, diminishing the remainder of 1234 by an amount x(as it is clear 183 cannot completely divide 1234). Hence, it will be a term (1234-x) which will be the remainder when N is divided by 183. It comes out that when 183 is multiplied by 6 the product is 1098 and when multiplied by 7 the product is 1281(greater than the term 1234 and of course when added to N-1234 yields a number greater than N which is not at par with the question.) Thus Subtracting 1098 from 1234 we yield a term 136 which is finally the remainder(i.e. can no more be divided by 183.)

Master, Kindly comment on this explanation and oblige. I will be so very grateful to you for pointing out my misunderstanding of the problem.

Tamoghna Banerjee
- 7 years, 9 months ago

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Yes, replacing $N$ with $N-1234$ will work.

The rest of the proof can be expressed clearer, but I understand what you are trying to say.

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Thank you Master, I will certainly strive for a clearer way the next time.

Tamoghna Banerjee
- 7 years, 9 months ago

The number N can be written in the format :

Dividend = Divisor * Quotient + Remainder as

N = 2013 * Q + 1234

This can be written as

N = 183 * 11 * Q + 1234

Dividing both sides by 183 , we get

N/183 = 11 * Q + 1234/183

N/183 = 11 * Q + (1098 +136)/183

N/183 = 11 * Q + 6 + 136/183

Multiplying both sides by 183 , we get

N = 183 * ( 11* Q + 6 ) + 136

If 11* Q + 6 = K

N = 183 * K + 136

which means 136 is the remainder when N is divided by 183

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N Mod 2013 = 1234,

N mod 183 = a,

N = 2013 + 1234 = 3247,

N mod 183 = 3247 Mod 183 = 136,

a = 136,

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N=2013a+1234=3.11.61.a+1234

N/183=1234/183=6.7431693989071038251366120218579

6.7431693989071038251366120218579-6= 0.7431693989071038251366120218579.183=136

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"If the integer N leaves a reminder of 1234 when divided by 2013"

since brilliant mathematics answers are integer values from 0 to 999 inclusive, so logically we can assume that N is 1234. N divided 2013 leaves 2013.

Now we find out what is the remainder when N is divided by 183, so :

$= 1234$ mod $183$

$= 136$

Thus it means the correct answer is 136

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That doesn't matter. 2013 is divisible by 183, this means that we know that we only need to deal with 1234. N could be any multiple of 2013. But yes, the rest of it is right.

Ryan Wood
- 7 years, 9 months ago

wew

Ronald Salim
- 7 years, 9 months ago

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The number N can be written in the format :

Dividend = Divisor * Quotient + Remainder as

N = 2013 * Q + 1234

This can be written as

N = 183 * 11 * Q + 1234

Dividing both sides by 183 , we get

N/183 = 11 * Q + 1234/183

N/183 = 11 * Q + (1098 +136)/183

N/183 = 11 * Q + 6 + 136/183

Multiplying both sides by 183 , we get

N = 183 * ( 11* Q + 6 ) + 136

If 11* Q + 6 = K

N = 183 * K + 136

which means 136 is the remainder when N is divided by 183