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?
Answer & explanation
Correct answer: option 3
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