Number System Test - 4

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.

Suppose the numbers are x and y, then:
x(x + y) = 3666 … (1)
y(x + y) = 2418 … (2)

Adding 1 and 2, we get
(x + y)² = 6084

Therefore, x + y = 78 … (3)

Subtracting (2) from (1), we get
x² - y² = 1248

Or, (x + y)(x - y) = 1248
Or, 78(x - y) = 1248
Or, x - y = 16

Try with some natural numbers and get the answer.

x / ( 3/8 ) * 5/6 = x* 8/3 *5/6 = x*40/18
equivalent to dividing by 18/40 = 9/20

N = 238a + 79 =17 × 14a + 17 × 4 + 11.
N = 17(14a + 4) +11

Hence, on dividing N by 17, the remainder is 11.

17 when divided by 16 leaves 1 as remainder and we have 23 such 17s which when divided by 16 leave 1 as remainder.