Mathematics

Use Euclid's division lemma to show that the cube of any positive integer is of form $$7m$$ or $$7m+1$$ or $$7m+6$$.


SOLUTION
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 $$0 \le r < b$$ 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
$${ \left( a+b \right)  }^{ 3 }=\quad { a }^{ 3 }+3{ a }^{ 2 }b+3a{ b }^{ 2 }+{ b }^{ 3 }$$
In this formula value of a is always $$7q$$
so collect the value we get
$${ \left( 7q+b \right)  }^{ 3 }=\quad 343q^{ 3 }+{ 147q }^{ 2 }b+21q{ b }^{ 2 }+{ b }^{ 3 }$$
Now divide by $$7$$ , we get  quotient = $$49q^{ 3 }+{ 21q }^{ 2 }r+3q{ r }^{ 2 }$$ 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 $$7m+6$$
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