Number System Test - 3

Considering the unit digit only in number like 509, 519, 529,...,589, 9 occurs 9 times .
Considering the ten digit from 590 - 598, (590,591,592,...,598) 9 occurs 9 times.
Hence, there are 9 + 9 = 18 numbers between 500 to 600 that contain the digit nine exactly.

All the number in this series except 2! are divisible by 3, therefore, reminder will be 2.

If n is even, then even + even + 1 = odd
If n is odd, then odd + odd + 1 = even
Therfore, n to the power 4 + n to the power 2 + 1 is always odd.

D = abcd / 9999, 49995 = 9999 * 5. Hence, once we multiply D with 9999 we will get a natural number.

The total number of multiples of 5 in 30! = 6 (5, 10, ….., 30)
The total number of multiples of 5 to the power 2 = 25 in 30! = 1 (25 only)
Further powers of 5 cannot be these as 5 to the power 3 = 125 > 30
Therefore, the greatest power of 5 that divides 30! exactly = 6 + 1 = 7

Suppose number are 2n, 2n + 2 and 2n + 4, where n is a whole number 2n (2n + 2) (2n + 4) = 2*2*2 (n) (n + 1) (n + 2) .
Now, n (n + 1) (n + 2) is the product of three consecutive natural number and at least one of them will be divisble by 2 and at least one of them will be divisible by three.

Therefore, n (n + 1) (n + 2) is divisible by 6.
Therefore, 8 (n) (n + 1) (n + 2) will be always be divisble by 48.

Choices (b), (c) and (d) will have a factor of 5 in the LCM (they have multiple of 5 in number). So, the only logical choice is (a).

From 259 to 458, there are two hundred natural numbers, so there will be 2 x 20 = 40 8s.
From 459 to 492, we have 13 more 8s; so the answer is 40 + 13 = 53.