Question

Find the least number which when divided by 8,12 and 20 leaves in each case a remainder 1, but when divided by 13 leaves no remainder.

Answer 1

LCM of 8, 12, 20 is :  120
   Number N =  13 * n = 120 m * + 1

  120 = 13 * 9 + 3
 => 13 * n = 13 * 9 * m + 3 * m + 1

To find the least value of N,  let 13 =  3 * m + 1
       =>  3m = 12
         so  m = 4

Then,  N = 481 = 37 * 13  =  120 * 4 + 1

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To find the next numbers,
     make  3 m + 1 = 26 / 39 / 52 / 65 / 78 / 91 / 104/ 117 / 130 / 143 / 156 / 169
 
Then  m = 17, 30, 43 , 56 ....

Thus the numbers will be :  481, 2,041 , 3601, ....