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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Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
Enrolled Students 87

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