Let x be the least number which when divided by 16,24,30,36 and 45, the remainder in each case is 4 and x is divisible by 28. If the HCF of x and 3193 is y, then what is the sum of the digits of y?

4

16, 24, 30, 36, 45 = 720 (LCM)

x = \(\frac{720n + 4}{28}\)

   = \(\frac{(720\;×\;4)\;+\;4}{28}\)     (Divided)

Hence,

x = 2880 + 4 = 2884               

Now, 2884, 3193

HCF of 2884, 3193 = y = 103

Sum of digits of y = 1 + 0 + 3 = 4