use Euclid division lemma show that the cube of any positive integer is of the form of 7m, 7m+1,7m+6
Answers
According to Euclid's Division Lemma
Let us take a as any positive integer and b=7
Then using Euclid's algorithm,
we get a=7q+r ;here r is remainder and value of q is more than or equal to 0 and r=0,1,2,3,4,5,6 because 0≤r<b and the value of b=7
So possible forms will 7q,7q+1,7q+2,7q+3,7q+4,7q+5,7q+6
to get the cube of these values use the formula
(a+b)
3
=a
3
+3a
2
b+3ab
2
+b
3
In this formula value of a is always 7q
so collect the value we get
(7q+b)
3
=343q
3
+147q
2
b+21qb
2
+b
3
Now divide by 7 , we get quotient = 49q
3
+21q
2
r+3qr
2
and remainder is b
3
so we have to consider the value of b
3
b=0 we get 7m+0=7m
b=1 then 1
3
=1 so we get 7m+1
b=2 then 2
3
=8 divided by 7 we get 1 as remainder so we get 7m+1
b=3 then 3
3
=27 divided by 7 we get 6 as remainder so we get 7m+6
b=4 then 4
3
=64 divided by 7 we get 1 as remainder so we get 7m+1
b=5 then 5
3
=125 divided by 7 we get 6 as remainder so we get 7m+6
b=6 then 6
3
=216 divided by 7 we get 6 as remainder so we get 7m+6
so all values are in the form of 7m,7m+1 and 7m+6