show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3
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Solution:
let n be any positive integer and b=3
n =3q+r
where q is the quotient and r is the remainder
0_ <r<3
so the remainders may be 0,1 and 2
so "n" may be in the form of 3q, 3q=1,3q+2
CASE-1
IF N=3q
n+4=3q+4
n+2=3q+2
here n is only divisible by 3
CASE-2
if n = 3q+1
n+4=3q+5
n+2=3q=3
here only n+2 is divisible by 3
CASE-3
IF n=3q+2
n+2=3q+4
n+4=3q+2+4
=3q+6
here only n+4 is divisible by 3
HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE
MARK BRAINLIEST!!
let n be any positive integer and b=3
n =3q+r
where q is the quotient and r is the remainder
0_ <r<3
so the remainders may be 0,1 and 2
so "n" may be in the form of 3q, 3q=1,3q+2
CASE-1
IF N=3q
n+4=3q+4
n+2=3q+2
here n is only divisible by 3
CASE-2
if n = 3q+1
n+4=3q+5
n+2=3q=3
here only n+2 is divisible by 3
CASE-3
IF n=3q+2
n+2=3q+4
n+4=3q+2+4
=3q+6
here only n+4 is divisible by 3
HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE
MARK BRAINLIEST!!
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