To find the highest power of 9 that divides the above product, it is required to find the sum of powers of all 3's in the expansion
Sum of powers of 3's $=\left[\dfrac{99}{3^1}\right] + \left[\dfrac{99}{3^2}\right] + \left[\dfrac{99}{3^3}\right] + \left[\dfrac{99}{3^4}\right]+...$
where [] is greatest integer.
Since 3 divides $b$ with a result of 2 and a remainder of 1, $b = 2×3 + 1 = 7$.
Since number 3 divides a with a result of $b$ (which we now know equals 7) and a remainder of 2, $ a = b×3 + 2 = 7×3 + 2 =$ 23.
Easy Number System Question - 24
Q24.
When a particular positive number is divided by 5, the remainder is 2. If the same number is divided by 6, the remainder is 1. If the difference between the quotients of division is 3, then find the number.