Commander 38 Posted March 24, 2017 Can you find a number such thatIf 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 0 Share this post Link to post Share on other sites

Commander 38 Posted March 25, 2017 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 ! 0 Share this post Link to post Share on other sites

Sriman Dutta 44 Posted March 25, 2017 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. 0 Share this post Link to post Share on other sites

Commander 38 Posted March 25, 2017 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. 0 Share this post Link to post Share on other sites

Daecon 191 Posted March 25, 2017 Is it a prime number? 0 Share this post Link to post Share on other sites

Sriman Dutta 44 Posted March 25, 2017 I can't get the method. 0 Share this post Link to post Share on other sites

Daecon 191 Posted March 25, 2017 (edited) 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 0 Share this post Link to post Share on other sites

Sriman Dutta 44 Posted March 25, 2017 What do you mean by x!+sigma y= z ? 0 Share this post Link to post Share on other sites

imatfaal 2481 Posted March 25, 2017 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 3 Share this post Link to post Share on other sites

Commander 38 Posted March 25, 2017 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. 0 Share this post Link to post Share on other sites

imatfaal 2481 Posted March 25, 2017 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 0 Share this post Link to post Share on other sites

Commander 38 Posted March 26, 2017 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. 0 Share this post Link to post Share on other sites

Sensei 675 Posted March 28, 2017 (edited) 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 0 Share this post Link to post Share on other sites