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Help! A remainder is chasing me


Best Answer normdeplume , 19 September 2007 - 07:23 PM

My answer is below, is this the answer you were looking for or is there a lower one?

Spoiler for solution



Just because your voice reaches halfway around the world doesn't mean you are wiser than when it reached only to the end of the bar. Go to the full post


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67 replies to this topic

#51 PhoenixTears

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Posted 04 January 2010 - 06:48 AM

Wow, I haven't posted on these forums in a long time...
Heh, I was about to go all number theory and brute force the answer, but then I saw the smart solutions people were putting up, and went... 'wow.'
I'll answer the one that was answered less often? (But still answered?)

Another variation of the above problem is as follows:

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 1.
When I divide it by 4, the remainder is 1.
When I divide it by 5, the remainder is 1.
When I divide it by 6, the remainder is 1.
When I divide it by 7, the remainder is 1.
When I divide it by 8, the remainder is 1.
When I divide it by 9, the remainder is 1.
When I divide it by 10, the remainder is 1.

But when I divide it by 11, the remainder is 0 (just to make the procedure a little bit different).

Can you find it?


Spoiler for similar method...

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#52 Rising Force

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Posted 29 July 2010 - 12:27 PM

Great riddle, very close to the Chinese remainder theorem,

Here's my brute force code for original riddle:

int remainder = 1;
            int answer = 0;
            string result = string.Empty;

            for (int number = 1; number <= 10000; number++)
            {
                for (int divisor = 2; divisor <= 10; divisor++)
                {
                    if (number % divisor == remainder)
                    {
                        if (remainder == 9)
                        {
                            answer = number;
                            result += string.Format("{0} ", answer);
                            break;
                        }
                        remainder++;
                    }
                }
                remainder = 1;
            }

Code for the second riddle:

string result2 = string.Empty;
            for (int number = 11; number <= 100000; number++)
            {
                for (int divisor = 2; divisor <= 10; divisor++)
                {
                    if (number % divisor == 1)
                    {
                        if (divisor == 10)
                        {
                            ++divisor;
                            if (number % divisor == 0)
                            {
                                result2 += string.Format("{0} ", number);
                                break;
                            }
                        }
                    }
                    else break;
                }
            }

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#53 kasarekar

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Posted 07 July 2011 - 08:01 AM

There is a perfect solution.
Study the question inteligently. Assume that answer is X. If you add 1 to the answer X = Say X+1 this figur will be the exact divisible by each number from 1 to 10.
Therefore you need to find a LCM Lowest common Multiple of numbers 1 to 10. That is 2520.

Therefore Answer = 2519

Pramod Kasarekar

Edited by kasarekar, 07 July 2011 - 08:02 AM.

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#54 IlluVis

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Posted 08 October 2011 - 03:00 PM

Ohw, I thought it was much easier.

Is this possible?
x² - x

Because:

2² - 2 = 4 - 2 = 2 --> 2/2 = 1
3² - 3 = 9 - 3 = 6 --> 6/3 = 2
4² - 4 = 16 - 4 = 12 --> 12/4 = 3
5² - 5 = 25 - 5 = 20 --> 20/5 = 4
6² - 6 = 36 - 6 = 30 --> 30/6 = 5
7² - 7 = 49 - 7 = 42 --> 42/7 = 6
8² - 8 = 64 - 8 = 56 --> 56/8 = 7
9² - 9 = 81 - 9 = 72 --> 72/9 = 8
10² - 10 = 100 - 10 = 90 --> 90/10 = 9

You see?

Or wasn't that the question?
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#55 guppy

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Posted 08 October 2011 - 05:33 PM

no i think u have misunderstood the question... u answered it thinking that remainder means the quotient (the actual answer) we are asking for the remainder
for instance lets say u have 5/2 the answer would = to 2 and the remainder would be 1 (speaking in long division way and without continuing to turn it into a decimal)
so the question was what is the smallest number you can get that when u divide it by 2 your remainder is 1 and when u divide by 3 your remainder is 2 and so on and so on...
the answer is 2519
no one could find anyth smaller.(but clever job on your answer were that the objective :thumbsup:)hope that clears everything up :)

Edited by guppy, 08 October 2011 - 05:36 PM.

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#56 harikrishnan

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Posted 28 October 2011 - 05:41 PM

i have the best method

since all the remainders differ by 1 from the divisors so.
Spoiler for the best method

i can assure you no other method is better

Edited by harikrishnan, 28 October 2011 - 05:45 PM.

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#57 harikrishnan

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Posted 28 October 2011 - 05:45 PM

i have the best method

since all the remainders differ by 1 from the divisors so.
Spoiler for the best method

i can assure you no other method is better
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#58 Finix

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Posted 15 November 2011 - 09:52 AM

Love it! Great problem
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#59 krashlia

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Posted 06 December 2011 - 08:09 PM

29, i guess?
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#60 Swaraj

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Posted 15 December 2011 - 06:24 AM

Spoiler for the remainder

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