Question

I NEED help with a math problem.

Answer 1

Answer:

109

Step-by-step explanation:

You can brute-force this:

Candidate numbers that are divisable by 13 if you take away 5: 18 , 31 , 44 , 57 , 70 , 83 , 96 , 109 , 122 , 135 , 148

Candidate numbers that are divisable by 14 if you take away 11: 25 , 39 , 53 , 67 , 81 , 95 , 109 , 123 , 137

Only 109 is in both lists.

Answer 2

Answer:

109

Step-by-step explanation:

a) x - 5 = 13 x n      x = 13n + 5

b) x - 11 = 14 x m     x = 14m + 11

13n + 5 = 14m + 11

13n = 14m + 6

n = (14m + 6) / 13   means 14m + 6 is the multiple of 13

14m + 6    could be 13 , 26, 39, 52, 65, 78, 91, 104, 117

if m =                               1  ,  2     3   4    5    6   7

14m + 6 will be               20  34  48  62 76  90 104

m = 7

n = 8

x = 109