Question

Find m and n in n608m, so that the number is divisible by both 3 and 5.​

Answer 1

Answer:

The solutions are of the form(m,n)

Possible solutions are :

(0,1)

(0,4)

(0,7)

(5,2)

(5,5)

(5,8)

Step-by-step explanation:

The no. is n608m.

It is divisible by both 3 & 5.

Since it's divisible by 5 , m would be either 0 or 5.

Case 1: When m = 0

Now , sum of digits should be divisible by 3 , so that the whole no. is divisible by 3.

n+6+0+8+0 should be divisible by 3.

n+14 should be divisible by 3.

n can be 1 or 4 or 7.

Case 2 : When m = 5

n+6+0+8+5 should be divisible by 3.

19+n should be divisible by 3.

So , n can be 2 , 5 or 8.

Answer 2

Step-by-step explanation:

by divisibility rules the number is divisible by 5 whether once place has

0 or 5.

the sum of the digits of the number is divisible by 3 the number also divisible

..hence we can say possible (m,n) are..

(1, 0)

(2, 5)

(4, 0)

(5, 5)

(7, 0)

(8, 5)

n=2

m=5

n=1

m=0

n=4

m=0

n=5

m=5

n=7

m=0

n=8

m=5