Question

show that cube of any positive integer is 5m,5m+1,5m+3​

Answer 1

Answer:

Let a be any positive integer.

By Euclid's division lemma,

a = bm + r where b = 5

⇒ a = 5m + r

So, r can be any of 0, 1, 2, 3, 4

∴ a = 5m when r = 0, a = 5m + 1 when r = 1, a = 5m + 2 when r = 2, a = 5m + 3 when r = 3, a = 5m + 4 when r = 4

Case I : a = 5m

⇒ a2 = (5m)2 = 25m2

⇒ a2 = 5(5m2) = 5q, where q = 5m2

Case II : a = 5m + 1

⇒ a2 = (5m + 1)2 = 25m2 + 10 m + 1

⇒ a2 = 5 (5m2 + 2m) + 1 = 5q + 1, where q = 5m2 + 2m

Case III : a = 5m + 2

⇒ a2 = (5m + 2)2

⇒ a2 = 25m2 + 20m + 4

⇒ a2 = 5 (5m2 + 4m) + 4

⇒ a2 = 5q + 4 where q = 5m2 + 4m

Case IV: a = 5m + 3

⇒ a2 = (5m + 3)2 = 25m2 + 30m + 9

⇒ a2 = 5 (5m2 + 6m + 1) + 4

⇒ a2 = 5q + 4 where q = 5m2 + 6m + 1

Case V: a = 5m + 4

⇒ a2 = (5m + 4)2 = 25m2 + 40m + 16

⇒ a2 = 5 (5m2 + 8m + 3) + 1

⇒ a2 = 5q + 1 where q = 5m2 + 8m + 3

From all these cases, it is clear that square of any positive integer cannot be of the form 5q + 2 or 5q + 3.

Step-by-step explanation: Mark it as the brainliest!

Answer 2

Step-by-step explanation:

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