Use Euclid's division lemma to show that the cube of any positive integer is of the form of 9m, 9m+1 or 9m+8.
(Mathematics
Class 10th
Chapter real numbers
Exercise 1.1 5th problem)
Answers
Answered by
4
this is the answer of question no 5 of ex 1.1
Attachments:
sanjana309:
Why did u take b= 3...why not b=9?
we can take b = 9 , and give r = 0 , 1 , 2 , 3 , 4 etc...
But there will be wastage of time , by putting values of r as 0 , 1 etc ...
So for convenience , we have taken b = 3
hope this is clear to u
if u have any doubts feel free to contact me @deepika
Answered by
3
Hi there !!
Let "a" be any positive integer and
b = 3
BY Euclid's division lemma ;
a = 3q + r ,
where r = 0 , 1 , 2
When r = 0
a = 3q
a³ = (3q)³
= 27q³
= 9(3q³)
= 9 m , where m = 3q³
==============================
WHen r = 1
a = 3q + 1
a³ = (3q + 1)³
= 27q³ + 27q² +9q + 1
= 9(3q³ + 9q² + 3q ) + 1
= 9 m + 1 , where m = 3q³ + 9q² + 3q
===============================
When r = 2,
a = 3q + 2
a³ = (3q + 2)³
= 27 q³ + 54q² + 36q + 8
= 9 ( 3q³ +6q² + 4q)+8
= 9 m + 8 , where m = 3q³ +6q² + 4q
Hence it is clear that cube of any positive integer is of the form 9m , 9m + 1 or 9m +8.
Let "a" be any positive integer and
b = 3
BY Euclid's division lemma ;
a = 3q + r ,
where r = 0 , 1 , 2
When r = 0
a = 3q
a³ = (3q)³
= 27q³
= 9(3q³)
= 9 m , where m = 3q³
==============================
WHen r = 1
a = 3q + 1
a³ = (3q + 1)³
= 27q³ + 27q² +9q + 1
= 9(3q³ + 9q² + 3q ) + 1
= 9 m + 1 , where m = 3q³ + 9q² + 3q
===============================
When r = 2,
a = 3q + 2
a³ = (3q + 2)³
= 27 q³ + 54q² + 36q + 8
= 9 ( 3q³ +6q² + 4q)+8
= 9 m + 8 , where m = 3q³ +6q² + 4q
Hence it is clear that cube of any positive integer is of the form 9m , 9m + 1 or 9m +8.
Can you please tell me why b=3 but not 9?
Similar questions