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Use Euclids Division Lemma to Show Cube of Positive Integer Exercise 1.1 Chapter 1 Real Numbers Maths Class 10 NCERT

Use Euclids division lemma to show cube of positive integer, how to use Euclids division lemma, Chapter 1: Real Numbers Maths Class 10 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. Question 5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. is solved by our expert teachers. You can get ncert solutions and notes for class 10 chapter 1 absolutely free. NCERT Solutions for class 10 Maths Chapter 1: Real Numbers is very essencial for getting good marks in CBSE Board examinations

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Question 5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer:

According to Euclid’s Division Lemma
Let take a as any positive integer and b = 9. Then using Euclid’s algorithm we get a = 9q + 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 , 7 , 8, because 0 ≤r < b and the value of b is 9 Sp possible forms will  9q, 9q+1, 9q+2,9q+3,9q+4,9q+5,9q+6,9q+7 and 9q+8 to get the cube of these values use the formula (a+b)³ = a³ + 3a²b+ 3ab² + b³ In this formula value of a is always 9q
So plug the value we get (9q+b)³  = 729q³ + 243q²b + 27qb² + b³ Now divide by 9 we get quotient = 81q³ + 27q²b + 3qb² and remainder is b³ So we have to consider the value of b³ b = 0 we get 9m+0  = 9m b = 1 then 1³ = 1 so we get 9m +1 b = 2 then 2³ = 8 so we get 9m + 8 b = 3 then 3³ = 27 and it is divisible by 9 so we get 9m
b = 4 then 4³ =64 divide by 9 we get 1 as remainder so we get 9m +1 b=5 then 5³=125 divide by 9 we get 8 as remainder so we get 9m+8
b=6 then 6³=216 divide by 9 no remainder there so we get 9m
b=7 then 7³ = 343 divide by 9 we get 1 as remainder so we get 9m+1 b=8 then 8³ = 512 divide by 9 we get 8 as remainder so we get 9m+8
So all values are in form of 9m , 9m+1 or 9m+8

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