A natural number N when successively divided by 4,5 and 6 leaves remainder 2, 4 and 5 respectively. What is the sum of the remainders obtained when N is successively divided by 12 and 10?

Answer 1

Let N = another natural number + 2 = 4a + 2 = 5b + 4 = 6c + 5.
4a - 2 = 5b, so a - b = 3 (let's call this equation I).
4(a - b) = 20.
Now we have 4a + 2 = 5b + 20 (let's call this equation II).
So 5b - 5b - 20 = 2 - 4.
10 = 2c.
c = 5.
10 is divisible by both 10 and 12, so the sum is <<10+12=22>>22. Answer: \boxed{22}.