Answer to Question #176553 in Combinatorics | Number Theory for asd

Question

If n and b are positive integers greater than 1 and n is divisible by b, what

is the last digit of nb? Explain.

Answer 1

solution:-

given

 n and b are positive integers greater than 1 and n is divisible by b,

case - 1

let n = 2k₁ +1

k₁$\neq$ 0

this is odd number

b= 2k₂+1

k₂$\neq 0$

this odd number

as given n is divisible by b

and

if we multiply n and b

we get

nb = ( 2k₁+1 )(2k₂+1)

= it will come in the form of 2k +1 i.e odd number so it can end with 1,3,5,7,9

case -2

let n = 2k₁

k₁$\neq$ 0

this is even number

b= 2k₂+

k₂$\neq 0$

this is even number

as given n is divisible by b

and

if we multiply n and b

we get

nb = ( 2k₁)(2k₂)

this is also a even number

$\implies$ so the end digits of the number can be 2,4,6,8