Divisibility of 16 : Last four digit of a number is divisible by 16 then the number is divisible by 16.

So, 30M6 is divisible by 16

Put M= 1 , 3016 is not divisible by16.

M= 2 , 3026 is not divisible by16.

M= 3 , 3036 is not divisible by16.

M= 4 , 3046 is not divisible by16.

M= 5 , 3056 is divisible by16.

So, M = 5

Divisibility of 11 : - If the difference between sum of digits at even places and sum of digits at odd places is 0 or multiple of 11, then the number is divisible by 11.

34R0503056 is divisible by 11

(3 + R + 5 + 3 + 5) – (4 + 0 + 0 + 0 + 6) = 11

Or, 16 + R – 10 = 11

Or, R = 11 – 6 = 5

Value of R and M are 5 and 5 respectively.

Hence, option C is the correct answer.

New number = 763254 – 5(41) = 763049

Since, last digit of the new number is 9, therefore it should not be divisible by 2 and 5.

7 + 6 + 3 + 0 + 4 + 9 = 29, it is also not divisible by 3.

∴ The new number is divisible by 7.

Hence, option B is the correct answer.

Smallest two-digit numbers which is divisible by 6 = 12

Largest two-digit numbers which is divisible by 6 = 96

Number of two-digit numbers which is divisible by 6 =

=  = 14 + 1 = 15

LCM of 18, 24, 30, 40 and 42 = 2520

Required number = 2520n + 7

Now, 2520×4 + 7 = 10080 + 7

= 10087

Sum of the digits = 1+0+0+8+7 = 16

A number is divisible 8 and 11 both then the number must be divisible by 88.

Divisibility of 8: Last three digits of a number is divisible by 8 then the number is divisible by 8.

So, n48 is divisible by 8

Hence value of n will be 0,2,4,6 and 8

Smallest value of n =0 but 0 is not possible because n is a natural number.

n =2

Divisibility of 11: A number is divisible by 11 when difference between the sum of digits at even place and the sum of digits at odd place is either 0 or multiple of 11.

9m2365n48

(8+n+6+2+9) – (4+5+3+m) = 0 or multiple of 11

27-12-m = 0 or multiple of 11

15 -m = 11

m = 4

(m² × n²)= (4² × 2²)=64

150328

New number = 853210

Resultant number = 853210 – 5 × 13 = 853210 – 65 = 853145

Hence, 853145 is divisible by 5.

Hence, option B is the correct answer.

Divisibility of 9 : - Sum of digits of a number is multiple of 9 then the number is divisible by 9.

x + 5 + 8 + y = 13 + x + y

Next multiple of 9 greater than 13 is 18.

13 + x + y = 18

Or, x + y = 18 – 13 = 5

Hence, option B is the correct answer.

We know that, a number is divisible by 3 if the sum of digits of a number is divisible by 3.

∴ There is no effect on divisibility of 3 if the digits are rearranged.

Therefore, number will be divisible by 3.

Hence, option A is the correct answer.

Let the digits at hundredth and tenth places be x and y respectively.

Divisibility of 11: A number if divisible by 11 if the difference of sum of digits at odd places and sum of digits at even places is either divisible by 11 or 0.

(4 + 6 + y) – (7 + x) = 10 + y – 7 – x

= 3 – x + y (i)

Divisibility of 3: A number is divisible by 3 if the sum of digits of a number is divisible by 3.

4 + 7 + 6 + x + y + 0

= 17 + x + y (ii)

By hit and trial

x = 8 and y = 5 satisfies both the equations.

Hence, option D is the correct answer.

Now, 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45, which is divisible by 9.

∴ Remainder = 0

Hence, option B is the correct answer.