2. Prove that the product of two consecutive positive integers is divisible by 2.
Solution:
Let’s consider two consecutive positive integers as (n-1) and n. ∴ Their product = (n-1) n = n 2 – n And then, we know that any positive integer is of form 2q or 2q+1. (From Euclid’s division lemma for b = 2) So, when n = 2q We have, ⇒ n 2 – n = (2q) 2 – 2q ⇒ n 2 – n = 4q 2 -2q ⇒ n 2 – n = 2(2q 2 -q) Thus, n 2 – n is divisible by 2. Now, when n = 2q+1 We have, ⇒ n 2 – n = (2q+1) 2 – (2q-1) ⇒ n 2 – n = (4q 2 +4q+1 – 2q+1) ⇒ n 2 – n = (4q 2 +2q+2) ⇒ n 2 – n = 2(2q 2 +q+1) Thus, n 2 – n is divisible by 2 again. Hence, the product of two consecutive positive integers is divisible by 2.3. Prove that the product of three consecutive positive integers is divisible by 6.
Solution:
Let n be any positive integer. Thus, the three consecutive positive integers are n, n+1 and n+2. We know that any positive integer can be of form 6q, or 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6). So, For n= 6q, ⇒ n(n+1)(n+2) = 6q(6q+1)(6q+2) ⇒ n(n+1)(n+2) = 6[q(6q+1)(6q+2)] ⇒ n(n+1)(n+2) = 6m, which is divisible by 6. [m = q(6q+1)(6q+2)] For n = 6q+1, ⇒ n(n+1)(n+2) = (6q+1)(6q+2)(6q+3) ⇒ n(n+1)(n+2) = 6[(6q+1)(3q+1)(2q+1)] ⇒ n(n+1)(n+2) = 6m, which is divisible by 6. [m = (6q+1)(3q+1)(2q+1)] For n = 6q+2, ⇒ n(n+1)(n+2) = (6q+2)(6q+3)(6q+4) ⇒ n(n+1)(n+2) = 6[(3q+1)(2q+1)(6q+4)] ⇒ n(n+1)(n+2) = 6m, which is divisible by 6. [m = (3q+1)(2q+1)(6q+4)] For n = 6q+3, ⇒ n(n+1)(n+2) = (6q+3)(6q+4)(6q+5) ⇒ n(n+1)(n+2) = 6[(2q+1)(3q+2)(6q+5)] ⇒ n(n+1)(n+2) = 6m, which is divisible by 6. [m = (2q+1)(3q+2)(6q+5)] For n = 6q+4, ⇒ n(n+1)(n+2) = (6q+4)(6q+5)(6q+6) ⇒ n(n+1)(n+2) = 6[(3q+2)(3q+1)(2q+2)] ⇒ n(n+1)(n+2) = 6m, which is divisible by 6. [m = (3q+2)(3q+1)(2q+2)] For n= 6q+5, ⇒ n(n+1)(n+2) = (6q+5)(6q+6)(6q+7) ⇒ n(n+1)(n+2) = 6[(6q+5)(q+1)(6q+7)] ⇒ n(n+1)(n+2) = 6m, which is divisible by 6. [m = (6q+5)(q+1)(6q+7)] Hence, the product of three consecutive positive integers is divisible by 6.4. For any positive integer n, prove that n 3 – n is divisible by 6.
Solution:
Let n be any positive integer. And any positive integer can be of form 6q, or 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6) We have n 3 – n = n(n 2 -1)= (n-1)n(n+1), For n = 6q, ⇒ (n-1)n(n+1) = (6q-1)(6q)(6q+1) ⇒ (n-1)n(n+1) = 6[(6q-1)q(6q+1)] ⇒ (n-1)n(n+1) = 6m, which is divisible by 6. [m = (6q-1)q(6q+1)] For n = 6q+1, ⇒ (n-1)n(n+1) = (6q)(6q+1)(6q+2) ⇒ (n-1)n(n+1) = 6[q(6q+1)(6q+2)] ⇒ (n-1)n(n+1) = 6m, which is divisible by 6. [m = q(6q+1)(6q+2)] For n = 6q+2, ⇒ (n-1)n(n+1) = (6q+1)(6q+2)(6q+3) ⇒ (n-1)n(n+1) = 6[(6q+1)(3q+1)(2q+1)] ⇒ (n-1)n(n+1) = 6m, which is divisible by 6. [m = (6q+1)(3q+1)(2q+1)] For n = 6q+3, ⇒ (n-1)n(n+1) = (6q+2)(6q+3)(6q+4) ⇒ (n-1)n(n+1) = 6[(3q+1)(2q+1)(6q+4)] ⇒ (n-1)n(n+1) = 6m, which is divisible by 6. [m = (3q+1)(2q+1)(6q+4)] For n = 6q+4, ⇒ (n-1)n(n+1) = (6q+3)(6q+4)(6q+5) ⇒ (n-1)n(n+1) = 6[(2q+1)(3q+2)(6q+5)] ⇒ (n-1)n(n+1) = 6m, which is divisible by 6. [m = (2q+1)(3q+2)(6q+5)] For n= 6q+5, ⇒ (n-1)n(n+1) = (6q+4)(6q+5)(6q+6) ⇒ (n-1)n(n+1) = 6[(6q+4)(6q+5)(q+1)] ⇒ (n-1)n(n+1) = 6m, which is divisible by 6. [m = (6q+4)(6q+5)(q+1)] Hence, for any positive integer n, n 3 – n is divisible by 6.5. Prove that if a positive integer is of form 6q + 5, then it is of form 3q + 2 for some integer q, but not conversely .
Solution:
Let n = 6q+5 be a positive integer for some integer q. We know that any positive integer can be of form 3k, or 3k+1, or 3k+2. ∴ q can be 3k or, 3k+1 or, 3k+2. If q = 3k, then ⇒ n = 6q+5 ⇒ n = 6(3k)+5 ⇒ n = 18k+5 = (18k+3)+ 2 ⇒ n = 3(6k+1)+2 ⇒ n = 3m+2, where m is some integer. If q = 3k+1, then ⇒ n = 6q+5 ⇒ n = 6(3k+1)+5 ⇒ n = 18k+6+5 = (18k+9)+ 2 ⇒ n = 3(6k+3)+2 ⇒ n = 3m+2, where m is some integer. If q = 3k+2, then ⇒ n = 6q+5 ⇒ n = 6(3k+2)+5 ⇒ n = 18k+12+5 = (18k+15)+ 2 ⇒ n = 3(6k+5)+2 ⇒ n = 3m+2, where m is some integer. Hence, if a positive integer is of form 6q + 5, then it is of form 3q + 2 for some integer q. Conversely, Let n = 3q+2 And we know that a positive integer can be of form 6k, or 6k+1, 6k+2, 6k+3, 6k+4, or 6k+5. So, if q = 6k+1, then ⇒ n = 3q+2 ⇒ n = 3(6k+1)+2 ⇒ n = 18k + 5 ⇒ n = 6m+5, where m is some integer. So, if q = 6k+2, then ⇒ n = 3q+2 ⇒ n = 3(6k+2)+2 ⇒ n = 18k + 6 +2 = 18k+8 ⇒ n = 6 (3k + 1) + 2 ⇒ n = 6m+2, where m is some integer Now, this is not of form 6q + 5. Therefore, if n is of form 3q + 2, then it is necessary won’t be of form 6q + 5.6. Prove that the square of any positive integer of form 5q + 1 is of the same form.
Solution:
Here, the integer ‘n’ is of form 5q+1. ⇒ n = 5q+1 On squaring it, ⇒ n 2 = (5q+1) 2 ⇒ n 2 = (25q 2 +10q+1) ⇒ n 2 = 5(5q 2 +2q)+1 ⇒ n 2 = 5m+1, where m is some integer. [For m = 5q 2 +2q]. Therefore, the square of any positive integer of form 5q + 1 is of the same form.7. Prove that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.
Solution:
Let any positive integer ‘n’ be of form 3q or 3q+1 or 3q+2. (From Euclid’s division lemma for b = 3) If n = 3q, Then, on squaring ⇒ n 2 = (3q) 2 = 9q 2 ⇒ n 2 = 3(3q 2 ) ⇒ n 2 = 3m, where m is some integer [m = 3q 2 ] If n = 3q+1, Then, on squaring ⇒ n 2 = (3q+1) 2 = 9q 2 + 6q + 1 ⇒ n 2 = 3(3q 2 +2q) + 1 ⇒ n 2 = 3m + 1, where m is some integer [m = 3q 2 +2q]. If n = 3q+2, Then, on squaring ⇒ n 2 = (3q+2) 2 = 9q 2 + 12q + 4 ⇒ n 2 = 3(3q 2 + 4q + 1) + 1 ⇒ n 2 = 3m + 1, where m is some integer [m = 3q 2 + 4q + 1]. Thus, it is observed that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.8. Prove that the square of any positive integer is of form 4q or 4q + 1 for some integer q.
Solution:
Let ‘a’ be any positive integer. Then, According to Euclid’s division lemma, a = bq+r According to the question, when b = 4. a = 4k + r, n < r < 4 When r = 0, we get, a = 4k a 2 = 16k 2 = 4(4k 2 ) = 4q, where q = 4k 2 When r = 1, we get, a = 4k + 1 a 2 = (4k + 1) 2 = 16k 2 + 1 + 8k = 4(4k + 2) + 1 = 4q + 1, where q = k(4k + 2) When r = 2, we get, a = 4k + 2 a 2 = (4k + 2) 2 = 16k 2 + 4 + 16k = 4(4k 2 + 4k + 1) = 4q, where q = 4k 2 + 4k + 1 When r = 3, we get, a = 4k + 3 a 2 = (4k + 3) 2 = 16k 2 + 9 + 24k = 4(4k 2 + 6k + 2) + 1 = 4q + 1, where q = 4k 2 + 6k + 2 Therefore, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.9. Prove that the square of any positive integer is of form 5q or 5q + 1, 5q + 4 for some integer q.
Solution:
Let ‘a’ be any positive integer. Then, According to Euclid’s division lemma, a = bq+r According to the question, when b = 5. a = 5k + r, n < r < 5 When r = 0, we get a = 5k a 2 = 25k 2 = 5(5k 2 ) = 5q, where q = 5k 2 When r = 1, we get a = 5k + 1 a 2 = (5k + 1) 2 = 25k 2 + 1 + 10k = 5k(5k + 2) + 1 = 5q + 1, where q = k(5k + 2) When r = 2, we get a = 5k + 2 a 2 = (5k + 2) 2 = 25k 2 + 4 + 20k = 5(5k 2 + 4k) + 4 = 4q + 4, where q = 5k 2 + 4k When r = 3, we get a = 5k + 3 a 2 = (5k + 3) 2 = 25k 2 + 9 + 30k = 5(5k 2 + 6k + 1) + 4 = 5q + 4, where q = 5k 2 + 6k + 1 When r = 4, we get a = 5k + 4 a 2 = (5k + 4) 2 = 25k 2 + 16 + 40k = 5(5k 2 + 8k + 3) + 1 = 5q + 1, where q = 5k 2 + 8k + 3 Therefore, the square of any positive integer is of form 5q or 5q + 1 or 5q + 4 for some integer q.10. Show that the square of an odd integer is of form 8q + 1 for some integer q.
Solution:
From Euclid’s division lemma, a = bq+r ; where 0 < r < b Putting b = 4 for the question, ⇒ a = 4q + r, 0 < r < 4 For r = 0, we get a = 4q, which is an even number. For r = 1, we get a = 4q + 1, which is an odd number. On squaring, ⇒ a 2 = (4q + 1) 2 = 16q 2 + 1 + 8q = 8(2q 2 + q) + 1 = 8m + 1, where m = 2q 2 + q For r = 2, we get a = 4q + 2 = 2(2q + 1), which is an even number. For r = 3, we get a = 4q + 3, which is an odd number. On squaring, ⇒ a 2 = (4q + 3) 2 = 16q 2 + 9 + 24q = 8(2q 2 + 3q + 1) + 1 = 8m + 1, where m = 2q 2 + 3q + 1 Thus, the square of an odd integer is of form 8q + 1 for some integer q.11. Show that any positive odd integer is of the form 6q +1 or 6q + 3 or 6q + 5, where q is some integer.
Solution:
Let ‘a’ be any positive integer. Then from Euclid’s division lemma, a = bq+r ; where 0 < r < b Putting b = 6, we get ⇒ a = 6q + r, 0 < r < 6 For r = 0, we get a = 6q = 2(3q) = 2m, which is an even number. [m = 3q] For r = 1, we get a = 6q + 1 = 2(3q) + 1 = 2m + 1, which is an odd number. [m = 3q] For r = 2, we get a = 6q + 2 = 2(3q + 1) = 2m, which is an even number. [m = 3q + 1] For r = 3, we get a = 6q + 3 = 2(3q + 1) + 1 = 2m + 1, which is an odd number. [m = 3q + 1] For r = 4, we get a = 6q + 4 = 2(3q + 2) + 1 = 2m + 1, which is an even number. [m = 3q + 2] For r = 5, we get a = 6q + 5 = 2(3q + 2) + 1 = 2m + 1, which is an odd number. [m = 3q + 2] Thus, from the above, it can be seen that any positive odd integer can be of the form 6q +1 or 6q + 3 or 6q + 5, where q is some integer.12. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.
Solution:
Let the positive integer = a According to Euclid’s division algorithm, a = 6q + r, where 0 ≤ r < 6 a 2 = (6q + r) 2 = 36q 2 + r 2 + 12qr [∵(a+b) 2 = a 2 + 2ab + b 2 ] a 2 = 6(6q 2 + 2qr) + r 2 …(i), where,0 ≤ r < 6 When r = 0, substituting r = 0 in Eq.(i), we get a 2 = 6 (6q 2 ) = 6m, where, m = 6q 2 is an integer. When r = 1, substituting r = 1 in Eq.(i), we get a 2 = 6 (6q 2 + 2q) + 1 = 6m + 1, where, m = (6q 2 + 2q) is an integer. When r = 2, substituting r = 2 in Eq(i), we get a 2 = 6(6q 2 + 4q) + 4 = 6m + 4, where, m = (6q 2 + 4q) is an integer. When r = 3, substituting r = 3 in Eq.(i), we get a 2 = 6(6q 2 + 6q) + 9 = 6(6q 2 + 6q) + 6 + 3 a 2 = 6(6q 2 + 6q + 1) + 3 = 6m + 3, where, m = (6q + 6q + 1) is integer. When r = 4, substituting r = 4 in Eq.(i), we get a 2 = 6(6q 2 + 8q) + 16 = 6(6q 2 + 8q) + 12 + 4 ⇒ a 2 = 6(6q 2 + 8q + 2) + 4 = 6m + 4, where, m = (6q 2 + 8q + 2) is integer. When r = 5, substituting r = 5 in Eq.(i), we get a 2 = 6 (6q 2 + 10q) + 25 = 6(6q 2 + 10q) + 24 + 1 a 2 = 6(6q 2 + 10q + 4) + 1 = 6m + 1, where, m = (6q 2 + 10q + 4) is integer. Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m. Hence Proved.13. Show that the cube of a positive integer of form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.
Solution:
Given, 6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5 Then, the positive integers are of form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5. Taking cube on L.H.S and R.H.S, For 6q, (6q) 3 = 216 q 3 = 6(36q) 3 + 0 = 6m + 0, (where m is an integer = (36q) 3 ) For 6q+1, (6q+1) 3 = 216q 3 + 108q 2 + 18q + 1 = 6(36q 3 + 18q 2 + 3q) + 1 = 6m + 1, (where m is an integer = 36q 3 + 18q 2 + 3q) For 6q+2, (6q+2) 3 = 216q 3 + 216q 2 + 72q + 8 = 6(36q 3 + 36q 2 + 12q + 1) +2 = 6m + 2, (where m is an integer = 36q 3 + 36q 2 + 12q + 1) For 6q+3, (6q+3) 3 = 216q 3 + 324q 2 + 162q + 27 = 6(36q 3 + 54q 2 + 27q + 4) + 3 = 6m + 3, (where m is an integer = 36q 3 + 54q 2 + 27q + 4) For 6q+4, (6q+4) 3 = 216q 3 + 432q 2 + 288q + 64 = 6(36q 3 + 72q 2 + 48q + 10) + 4 = 6m + 4, (where m is an integer = 36q 3 + 72q 2 + 48q + 10) For 6q+5, (6q+5) 3 = 216q 3 + 540q 2 + 450q + 125 = 6(36q 3 + 90q 2 + 75q + 20) + 5 = 6m + 5, (where m is an integer = 36q 3 + 90q 2 + 75q + 20) Hence, the cube of a positive integer of form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.14. Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.
Solution:
According to Euclid’s division Lemma, Let the positive integer = n And, b=5 n = 5q+r, where q is the quotient and r is the remainder. 0 < r < 5 implies that the remainder may be 0, 1, 2, 3, 4 and 5. Therefore, n may be in the form of 5q, 5q+1, 5q+2, 5q+3, and 5q+4. So, this gives us the following cases: CASE 1: When, n = 5q n+4 = 5q+4 n+8 = 5q+8 n+12 = 5q+12 n+16 = 5q+16 Here, n is only divisible by 5. CASE 2: When, n = 5q+1 n+4 = 5q+5 = 5(q+1) n+8 = 5q+9 n+12 = 5q+13 n+16 = 5q+17 Here, n + 4 is only divisible by 5. CASE 3: When, n = 5q+2 n+4 = 5q+6 n+8 = 5q+10 = 5(q+2) n+12 = 5q+14 n+16 = 5q+18 Here, n + 8 is only divisible by 5. CASE 4: When, n = 5q+3 n+4 = 5q+7 n+8 = 5q+11 n+12 = 5q+15 = 5(q+3) n+16 = 5q+19 Here, n + 12 is only divisible by 5 CASE 5: When, n = 5q+4 n+4 = 5q+8 n+8 = 5q+12 n+12 = 5q+16 n+16 = 5q+20 = 5(q+4) Here, n + 16 is only divisible by 5. So, we can conclude that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5. Hence Proved.15. Show that the square of an odd integer can be of form 6q + 1 or 6q + 3 for some integer q.
Solution:
Let ‘a’ be an odd integer and b = 6. According to Euclid’s algorithm, a = 6m + r for some integer m ≥ 0 And r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6. So, we get, a = 6m or, 6m + 1 or, 6m + 2 or, 6m + 3 or, 6m + 4 or 6m + 5 Thus, we are choosing for a = 6m + 1 or, 6m + 3 or 6m + 5 for it to be an odd integer. For a = 6m + 1, (6m + 1) 2 = 36m 2 + 12m + 1 = 6(6m 2 + 2m) + 1 = 6q + 1, where q is some integer and q = 6m 2 + 2m. For a = 6m + 3 (6m + 3) 2 = 36m 2 + 36m + 9 = 6(6m 2 + 6m + 1) + 3 = 6q + 3, where q is some integer and q = 6m 2 + 6m + 1 For a = 6m + 5, (6m + 5) 2 = 36m 2 + 60m + 25 = 6(6m 2 + 10m + 4) + 1 = 6q + 1, where q is some integer and q = 6m 2 + 10m + 4. Therefore, the square of an odd integer is of form 6q + 1 or 6q + 3 for some integer q. Hence Proved.16. A positive integer is of the form 3q + 1, q is a natural number. Can you write its square in any form other than 3m + 1, 3m or 3m + 2 for some integer m? Justify your answer.
Solution:
No, we cannot write its square in any form other than 3m + 1, 3m or 3m + 2 for some integer m. Justification: By Euclid’s Division Lemma, a = bq + r, 0 ≤ r < b Here, a is any positive integer and b = 3, ⇒ a = 3q + r So, a can be of form 3q, 3q + 1 or 3q + 2. Now, for a = 3q (3q) 2 = 3(3q 2 ) = 3m [where m = 3q 2 ] for a = 3q + 1 (3q + 1) 2 = 9q 2 + 6q + 1 = 3(3q 2 + 2q) + 1 = 3m + 1 [where m = 3q 2 + 2q] for a = 3q + 2 (3q + 2) 2 = 9q 2 + 12q + 4 = 9q 2 + 12q + 3 + 1 = 3(3q 2 + 4q + 1) + 1 = 3m + 1 [where m = 3q 2 + 4q + 1] Thus, the square of a positive integer of form 3q + 1 is always of the form 3m + 1 or 3m for some integer m.17. Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.
Solution:
Let the positive integer be ‘a’ According to Euclid’s division lemma, a = bm + r According to the question, we take b = 3 a = 3m + r So, r = 0, 1, 2. When r = 0, a = 3m. When r = 1, a = 3m + 1. When r = 2, a = 3m + 2. Now, When a = 3m a 2 = (3m) 2 = 9m 2 a 2 = 3(3m 2 ) = 3q, where q = 3m 2 When a = 3m + 1 a 2 = (3m + 1) 2 = 9m 2 + 6m + 1 a 2 = 3(3m 2 + 2m) + 1 = 3q + 1, where q = 3m 2 + 2m When a = 3m + 2 a 2 = (3m + 2) 2 a 2 = 9m 2 + 12m + 4 a 2 = 3(3m 2 + 4m + 1) + 1 a 2 = 3q + 1 where q = 3m 2 + 4m + 1 Therefore, the square of any positive integer cannot be of form 3q + 2, where q is a natural number. Hence Proved.
