A Very Simple Arithmetic Puzzle !

[Commander]

Can you find a number such that

If the number is divided by 3 , the remainder is 1

 

If the number is divided by 4 , the remainder is 2

 

If the number is divided by 5 , the remainder is 3

 

If the number is divided by 6 , the remainder is 4

 

If the number is divided by 7 , the remainder is 5

 

If the number is divided by 8 , the remainder is 6

 

If the number is divided by 9 , the remainder is 7

[Velocity_Boy]

418.

[Commander]

418.

 

Good try.

 

It works upto division by 7 but not for 8 & 9.

 

You are on the right track.

 

This is a very simple Puzzle which I had half a mind to edit and pull off.

 

I will continue to work and create a tougher puzzle !

[Sriman Dutta]

Do I need to find the answer by a mere guess work? Or is there some solution?

I framed it :

Let the number be n. Then n mod 3 = 1 or n=3q+ 1, q is quotient.

[Commander]

Do I need to find the answer by a mere guess work? Or is there some solution?

I framed it :

Let the number be n. Then n mod 3 = 1 or n=3q+ 1, q is quotient.

 

It can be found in both ways.

 

Not a guess work but a logical and progressive calculation.

 

Also there is a straight forward mathematical derivation too !

 

Goodluck.

[Daecon]

Is it a prime number?

[Sriman Dutta]

I can't get the method.

[Daecon]

I'm an idiot. If you divide by 4 with 2 left over, it has to be even and so can't be prime.

 

x(factoral)+(sigma)y=z?

Edited March 25, 2017 by Daecon

[Sriman Dutta]

What do you mean by x!+sigma y= z ?

[imatfaal]

 

22678mod2520 = 2518

 

 

I really would not call this very simple - unless you know Chinese Remainder or are very adept with modular arithmetic then it would strike me as very hard.

 

A few hints in the next spoiler

 

 

Everything which gives a remainder 7 when divided by 9 will give a remainder 1 when divided by 3 but not the other way around. You can dispense with a few of the terms to make the maths easier / possible

 

[Commander]

 

22678mod2520 = 2518

 

 

I really would not call this very simple - unless you know Chinese Remainder or are very adept with modular arithmetic then it would strike me as very hard.

 

A few hints in the next spoiler

 

 

Everything which gives a remainder 7 when divided by 9 will give a remainder 1 when divided by 3 but not the other way around. You can dispense with a few of the terms to make the maths easier / possible

 

 

Well done Imatfaal +1 for you !

 

What is 22678 composed of ?

 

I reached the solution 2518 in the following way.

 

2520 being the LCM of [3,4,5,6,7,8 & 9] 2520 will be divided by 3,4,5,6 etc without remainder.

 

2520-2 = 2518 will be divided by each number with the remainder of 3-2, 4-2, 5-2, 6-2 etc leaving the remainder 1, 2, 3, 4, etc.

[imatfaal]

 

Well done Imatfaal +1 for you !

 

What is 22678 composed of ?

 

I reached the solution 2518 in the following way.

 

2520 being the LCM of [3,4,5,6,7,8 & 9] 2520 will be divided by 3,4,5,6 etc without remainder.

 

2520-2 = 2518 will be divided by each number with the remainder of 3-2, 4-2, 5-2, 6-2 etc leaving the remainder 1, 2, 3, 4, etc.

 

Oh my word that is a much neater solution than mine. Well done.

 

I used Chinese remainder theorem which can be phrased thus

 

 

for a set of congruences which are simultaneous as follows:

 

[latex]x\equiv a_i(mod\ m_i)[/latex]

 

for

 

[latex] i=1,2,...k[/latex]

 

if one sets

 

[latex]M=m_1.m_2...m_r[/latex]

 

and

 

[latex]b_i \frac{M}{m_i}=1(mod\ m_i)[/latex]

 

then

 

[latex] x \equiv \left [ a_1.b_1 \frac{M}{m_1}+...+a_k.b_k\frac{M}{m_k}\right ](mod\ M)[/latex]

 

This only works if all pairs of modulos are pairwise coprime - but if that is the case it always works. As you can see yours is so much nicer and simpler. I did mine with the help of a spreadsheet for the arithmetic

[Commander]

 

Oh my word that is a much neater solution than mine. Well done.

 

I used Chinese remainder theorem which can be phrased thus

 

 

for a set of congruences which are simultaneous as follows:

 

[latex]x\equiv a_i(mod\ m_i)[/latex]

 

for

 

[latex] i=1,2,...k[/latex]

 

if one sets

 

[latex]M=m_1.m_2...m_r[/latex]

 

and

 

[latex]b_i \frac{M}{m_i}=1(mod\ m_i)[/latex]

 

then

 

[latex] x \equiv \left [ a_1.b_1 \frac{M}{m_1}+...+a_k.b_k\frac{M}{m_k}\right ](mod\ M)[/latex]

 

This only works if all pairs of modulos are pairwise coprime - but if that is the case it always works. As you can see yours is so much nicer and simpler. I did mine with the help of a spreadsheet for the arithmetic

 

☑ Thank you for your kind words and sharing the steps of your derivation.

 

The elegant Solution was possible as the difference between the divisor and the remainder is constant and in this case 2

 

A friend of mine who solved it first after I created this puzzle and posted it in a few groups used this elegant method after noting the constant difference.

[Sensei]

I really would not call this very simple - unless you know Chinese Remainder or are very adept with modular arithmetic then it would strike me as very hard.

 

For me it was very easy. You just have to know C/C++.

It took less than minute to make this:

#include <stdio.h>

/*
If the number is divided by 3 , the remainder is 1
If the number is divided by 4 , the remainder is 2
If the number is divided by 5 , the remainder is 3
If the number is divided by 6 , the remainder is 4
If the number is divided by 7 , the remainder is 5
If the number is divided by 8 , the remainder is 6
If the number is divided by 9 , the remainder is 7
*/

int main( int argc, int *argv[] )
{
    for( long long i = 0; i < 4e9; i++ )
    {
        if( ( i % 3 ) != 1 ) continue;
        if( ( i % 4 ) != 2 ) continue;
        if( ( i % 5 ) != 3 ) continue;
        if( ( i % 6 ) != 4 ) continue;
        if( ( i % 7 ) != 5 ) continue;
        if( ( i % 8 ) != 6 ) continue;
        if( ( i % 9 ) != 7 ) continue;
        printf( "i=%d\n", i );
        break;
    }
    return( 0 );
}

SimpleArithmeticPuzzle.zip

Edited March 29, 2017 by Sensei