1) Use Euclid's division lemma to show that the cube of any positive integer is of
the form 7m or 7m+1 or 7m+6.
Answers
Step-by-step explanation:
Given:-
Euclid's division lemma
To find:-
Use Euclid's division lemma to show that the cube of any positive integer is of the form 7m or 7m+1 or 7m+6.
Solution:-
We know that
Euclid's Division Lemma:-
For a pair of given positive integers ‘a’ and ‘b’ there exist unique
integers ‘q’ and ‘r’ such that
a = bq + r where 0 ≤ r < b.
Let a = 7q +r ,0≤r<7
We have r = 0,1,2,3,4,5,6
1) Put r = 0 then ,
a=7q+0
=>a = 7q
On cubing both sides
=>a^3 = (7q)^3
=>a^3 = 7×49q^3
=>a^3 = 7m-------(1)
Where m = 49 q^3
2) Put r = 1 then
a = 7q+1
On cubing both sides
=>a^3 = (7q+1)^3
=>a^3 = 343q^3+147q^2+21q+1
=>a^3 = 7(49q^3+21q^2+3q)+1
=>a^3 = 7m+1 ---------(2)
where m = 49q^3+21q^2+3q
3) Put r = 2 then
a = 7q+2
On cubing both sides
=>a^3 = (7q+2)^3
=>a^3=343q^3+294q^2+84q+8
=>a^3=343q^3+294q^2+84q+7+1
=>a^3=7(49q^3+42q^2+12q+1)+1
=>a^3=7m+1-------(3)
where m=49q^3+42q^2+12q+1
4)Put r=3 then a=7q+3
On cubing both sides then
=>a^3=(7q+3)^3
=>a^3=343q^3+441q^2+189q+27
=>a^3=7(49q^3+63q^2+27q+7)+6
=>a^3=7m+6----------(4)
where m=49q^3+63q^2+27q+7
5) Put r = 4 then a= 7q+4 then
On cubing both sides then
=>a^3=(7q+4)^3
=>a^3=343q^3+588q^2+336q+64
=>a^3=7(49q^3+84q^2+48q+9)+1
=>a^3=7m+1---------(5)
where m = 49q^3+84q^2+48q+9
6)Put r=5 then a = 7q+5
On cubing both sides then
=>a^3=(7q+5)^3
=>a^3 = 343q^3+735q^2+525q+125
=>a^3=7(49q^3+105q^2+75q+17)+6
=>a^3=7m+6--------(6)
where m=49q^3+105q^2+75q+17
7)Put r =6 then a = 7q+6
On cubing both sides then
=>a^3=(7q+6)^3
=>a^3 =343q^3+882q^2+756q+216
=>a^3=7(49q^3+126q^2+108q+30)+6
=>a^3=7m+6-------------(7)
wherem=49q^3+126q^2+108q+30
From above all equations we conclude that
the cube of any positive integer is of the form 7m or 7m+1 or 7m+6.
Answer:-
The cube of any positive integer is of
the form 7m or 7m+1 or 7m+6.
Used formulae:-
- Euclid's Division Lemma:-
For a pair of given positive integers ‘a’ and ‘b’ there exist unique integers ‘q’ and ‘r’ such that a = bq + r where 0 ≤ r < b.
- (a+b)^3 = a^3+3a^2b+3ab^2+b^3
Given : cube of any positive integer
To Find : cannot be in the form
1) 7m
2) 7m + 1
3) 7m + 3
4) 7m + 6
Solution:
Any positive can be represented in the form a = bq + r
Without losing generality Hence any number can be of the form :
7k , 7k + 1 , 7k + 2 , 7k + 3 , 7k + 4 , 7k + 5 , 7k + 6
Lets cube each number
(7k)³ = 7 * 49k³ = 7m
(a + b)³ = a³ + b³ + 3ab(a + b)
(7k + 1)³ = (7k)³ + 1 + 3*7k *(7k + 1) = 7k ( 49k² + 21k +3) + 1 = 7m + 1
(7k + 2)³ = (7k)³ + 8 + 3*7k*2 *(7k +3) = 7k ( 49k² + 42k + 18) + 7 + 1
= 7k ( 49k² + 42k + 18 + 1 ) + 1
= 7m + 1
(7k + 3)³ = (7k)³ + 27 + 3*7k*3 *(7k +3) = 7k ( 49k² + 63k + 27) + 21 + 6
= 7k ( 49k² + 63k + 27 + 31 ) + 6
= 7m + 6
Similarly (7k + 4)³ = 7m + 1
(7k + 5)³ = 7m + 6
(7k + 6)³ = 7m + 6
7m , 7m + 1, 7m + 6 are three forms of cube of any positive integer
Hence cube of any positive integer cannot be in the form 7m + 3 from the given options
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