Math, asked by harman96, 1 year ago

show that the cube of any positive integer is of the form 4m, 4m+1 or 4m+3 for some integer m

Answers

Answered by Kanikashah
21
Let a be the positive integer n 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.
(4q)3 = 64q3 = 4(16q3)
= 4m, where m is some integer.

(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.

(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.

Hope it helps ...!!!
Plz mark as brainliest...

harman96: thx
Kanikashah: Most wlc.....hope it helps...
harman96: yeah
harman96: it is right
Answered by Krinshusahu
6
this is answer of your question
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