Question

13. Show that cube of any positive integer is either of the form 4q,4q + 1 or 4q+ 3
NCERT Exemplar
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Answer 1

Answer:

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.

(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.

Answer 2

Answer:

Answer:

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.

(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❤❤❤❤❤❤❤❤❤❤