show that the cube of any positive integer is of the form 4m or 4m+1 or 4m+3 where‘m’ is a whole number
Answers
Step-by-step explanation:
Solution: Let a be the positive integer and b = 4. Then, by Euclid’s algorithm,
a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.
So,
a = 4q or 4q + 1 or 4q + 2 or 4q + 3.
(1) . (4q)3 = 64q3 = 4(16q3) = 4m,
where m is some integer.
(2). (4q + 1)3 = 64q3 + 48q2 + 12q + 1 = 4(16q3 + 12q2 + 3) + 1 = 4m + 1, where m is some integer.
(4q + 2)3 = 64q3 + 96q2 + 48q + 8 = 4(16q3 + 24q2 + 12q + 2) = 4m, where m is some integer.
(3). (4q + 3)3 = 64q3 + 144q2 + 108q + 27 = 4(16q3 + 36q2 + 27q + 6) + 3 = 4m + 3, where m is some integer.
Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.Read more on Sarthaks.com - https://www.sarthaks.com/12812/show-that-cube-of-any-positive-integer-is-of-the-form-4m-4m-1-or-4m-3-for-some-integer-m
Given : cube of any positive integer is of the form 4m or 4m+1 or 4m+3
To Find : Prove
Solution:
Without losing generality Any number can be represented by
4k , 4k+1 . 4k+ 2 , 4k+3
(4k)³ = 64k³ = 4(16k³) = 4m
(4k + 1)³ = 64k³ + 3.16k².1 + 3.4k.1 + 1
= = 64k³ + 48k² + 12k + 1
= 4(16k³ + 12k² + 3k) + 1
= 4m + 1
(4k + 2)³ = 64k³ + 3.16k².2 + 3.4k.2² + 8
= = 64k³ + 96k² + 48k + 8
= 4(16k³ + 24k² + 12k + 2)
= 4m
(4k + 3)³ = 64k³ + 3.16k².3 + 3.4k.3² + 27
= 64k³ + 144k² + 108k + 27
= 64k³ + 144k² + 108k + 24 + 3
= 4(16k³ + 36k² +27k + 6) +3
= 4m +3
cube of any positive integer is of the form 4m or 4m+1 or 4m+3
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