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Mohan is selling cookies at the economics fair. As he decides how to package the cookies, he finds that when he bags them in groups of 4, he has 3 left over. When he bags them in groups of 5, he has 2 left over. When he bags them in groups of 7, he has 4 left over. What is the least number of cookies that Mohan could have?

 

Thanks for the last one @CalTheGreat!

 Mar 25, 2020
edited by Guest  Mar 25, 2020
edited by Guest  Mar 25, 2020
 #1
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We need to find the LCM of 17, 6, and 2 and add one to it.

 

The LCM of 17, 6, and 2 is 102 (prime factorization) 

102+1=103.

 

Therefore, our k=103

 

Hope this helped!

 Mar 25, 2020
 #2
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i=0;j=0;m=0;t=0;a=(4, 5, 7);r= (3, 2, 4);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return


  /* CHINESE REMAINDER THEOREM + MODULAR MULTIPLICATIVE INVERSE */

 

OUTPUT = 140 m +  67, where m =0, 1, 2, 3.......etc.

The minimum number of cookies =67

 Mar 25, 2020

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