HCF and LCM

A number p is such that:
p = 5k1 + 4,
p = 6k1 + 5,
p = 7k1 + 6,
p = 8k1 + 7,
p = 9k1 + 8,

Find the smallest number that satisfy this condition



Answer:

p = 5k1 + 4, or p + 1 = 5(k1 + 1)
p = 6k1 + 5, or p + 1 = 6(k2 + 1)
p = 7k1 + 6, or p + 1 = 7(k3 + 1)
p = 8k1 + 7, or p + 1 = 8(k4 + 1)
p = 9k1 + 8, or p + 1 = 9(k5 + 1)

Since p + 1 is a multiple of 5, 6, 7, 8, 9 we find the LCM of 5, 6, 7, 8, 9.
i.e = 2520.
this is the smallest value of p + 1 that will satisfy the above equations.
So, p = 2520 - 1 = 2519.