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

Question #176553

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

is the last digit of nb? Explain.


1
Expert's answer
2021-04-29T18:06:13-0400

solution:-


given

 and are positive integers greater than 1 and 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


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Comments

Michael Collado
10.04.21, 18:34

Answer is 0, if you notice, any number n can be divisible by itself so b = n. if n = 2 base 10 converting it to base 2 (b) will result in 10 where last digit is 0. same thing happens if you try 3,5,7 if you try n = 16, possible values of b would be 2,4,8 and 16 converting 16 to those bases will give you 10, 20, 100 and 10000 respectively which all have 0 as last digit

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