# 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.

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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

====

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, ....

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

====

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, ....