3 Marks Question

Question 513 Marks

Show that the cube of a positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

Answer

Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that.
a = 6q + r, where, 0 ≤ r < 6
⇒ a³ = (6q + r)³ = 216q³ + r³ + 3.6q.r(6q + r)
[$\because$ (a + b)³ = a³ + b³ + 3ab(a + b)]
⇒ a³ = (216q³ + 108q²r + 18qr²) + r³ ...(i)
Where, 0 ≤ r < 6
Case I: When r = 0, then putting r = 0 in
Eq. (i), we get
a³ = 216q³ = 6(36q³) = 6m
Where, m = 36q³ is an integer.
Case II: Where r = 1, then putting r = 1 in
Eq. (i), we get
a³ = (216q³ + 108q³ + 18q) + 1 = 6(36q³ + 18q³ + 3q) + 1
a³ = 6m + 1,
Where m = (36q³ + 18q³ + 3q) is an integer.
Case III: When r = 2, then putting r = 2 in
Eq. (i), we get
a³ = (216q³ + 216q² + 72q) + 8
a³ = (216q³ + 216q² + 72q + 6) + 2
⇒ a³ = 6(36q³ + 36q² + 12q + 1) + 2= 6m + 2
Where, m = (36q³ + 36q² + 12q + 1) is an integer.
Case IV: When r = 3, then putting r = 3 in
Eq. (i), we get
a³ = (216q³ + 324q² + 162q) + 27
= (216q³ + 324q² + 162q + 24) + 3
= 6(36q³ + 54q² + 27q + 4) + 3 = 6m + 3
Where, m = (36q³ + 64q² + 27q + 4) is an integer.
Case V: When r = 4, then putting r = 4 in
Eq. (i), we get
a³ = (216q³ + 432q² + 288q) + 64
a³ = 6(36q³ + 72q² + 48q) + 60 + 4
a³ = 6(36q³ + 72q² + 48q + 10) is an integer.
Case VI: When r = 5, then putting r = 5 in
Eq. (i), we get
a³ = (216q³ + 540q² + 450q) + 125
⇒ a³ = (216q³ + 540q² + 450q) + 120 + 5
⇒ a³ = 6(36q³ + 90q² + 75q + 20) + 5
⇒ a³ = 6m + 5
Where, m = (36q³ + 90q² + 75q + 20) is an integer.
Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the forms 6m, 6m + 1, 6m + 3, 6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.

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Question 523 Marks

Prove that the product of three consecutive positive integer is divisible by 6.

Answer

Let, n be any positive integer. Since any positive is of the form 6q or 6q + 1 or, 6q + 2 or, 6q + 3 or 6q + 4 or 6q + 5.
If n - 6q, then

n(n + 1)(n + 2) = [(6q)(6q + 1)(6q + 2)]

= 6[q(6q + 1)(6q + 2)]

= 6m, which is divisible by 6

If n - 6q + 1, then

n(n + 1)(n + 2) = (6q + 1)(6q + 2)(6q + 3)

= 6[(6q + 1)(3q + 1)(2q + 1)]

= 6m, which is divisible by 6

If n - 6q + 2, then

n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)

= 6[(3q + 1)(2q + 1)(6q + 4)]

= 6m, which is divisible by 6

If n - 6q + 3, then

n(n + 1)(n + 2) = (6q + 3)(6q + 4)(6q + 5)

= 6[(3q + 1)(3q + 2)(6q + 5)]

= 6m, which is divisible by 6

If n - 6q + 4, then

n(n + 1)(n + 2) = (6q + 4)(6q + 5)(6q + 6)

= 6[(6q + 4)(6q + 5)(q + 1)]

= 6m, which is divisible by 6

If n - 6q + 5, then

n(n + 1)(n + 2) = (6q + 5)(6q + 6)(6q + 7)

= 6[(6q + 5)(q + 1)(6q + 7)]

= 6m, which is divisible by 6

Hence, the product of three consecutive positive integer is divisible by 6.

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Maths 3 Marks Question