66 replies to this topic

[PhoenixTears]

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...

Like some people have said, the solution is probably only really able to found through brute force... kind of.
Use the LCM method, and you find that the number x-1 is divisible by 2, 3, 4, 5, 6, 7, 8, 9, and 10, where x will be the number that fits these conditions.
x-1 = some multiple of 2520. Divide 2520 by 11, and you get a remainder of 1. You figure out, you need to multiply the number by ten, because you will be adding one, making the remainder 11, or 0.
Hence, 25201.

[Rising Force]

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;
                }
            }

[kasarekar]

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.

[IlluVis]

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?

[guppy]

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 )hope that clears everything up

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

[harikrishnan]

i have the best method

since all the remainders differ by 1 from the divisors so.

Spoiler for the best method

.. let us assume the number to by x
so adding one to x ie x+1 results another number of special property.
ie it is divisible by all other numbers 2,3,4,5,6,7,8,9,10.
to find the least number taking their LCM(lowest common multiple) results you 2520 yes
x+1=2520
so x is equal to 2519

i can assure you no other method is better

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

[harikrishnan]

i have the best method

since all the remainders differ by 1 from the divisors so.

Spoiler for the best method

.. let us assume the number to by x
so adding one to x ie x+1 results another number of special property.
ie it is divisible by all other numbers 2,3,4,5,6,7,8,9,10.
to find the least number taking their LCM(lowest common multiple) results you 2520 yes
x+1=2520
so x is equal to 2519

i can assure you no other method is better

[Finix]

Love it! Great problem

[krashlia]

29, i guess?

[Swaraj]

Spoiler for the remainder

2519
Lets say the number is X.
SInce, it is divisible by the numbers 2,3,...,10, if we add 1 to it.
SO, the number (X+1) is divisible by 2,3,4,...,10.
Now, we need to find the smallest number divisible by 2,3,4,...,10.
It should be the LCM(2,3,4,...,10) = 2520
So, X+1 = 2520
hence, X = 2519