Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1 to get the best methods of solving problems.
TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.1
Question 1.
Use Euclid’s algorithm to find the HCF of
(i) 900 and 270
Answer:
900 and 270
Euclid Division Lemma
a = bq + r, q > 0 and 0 ≤ r < b
When 900 is divided 270, the remainder is 90 to get 900 = 270 × 3 + 90
Now consider the division of 270 with the remainder 90 in the above and division algorithm to get 270 = 90 × 3 + 0 The remainder is zero, when we cannot proceed further. We conclude that the HCF of (900, 270) = 90
(ii) 196 and 38220
Answer:
196 and 38220
When 38220 is divided 196, the remainder is 0 to get 38220 = 196 × 195 + 0
The remainder is zero, when we cannot proceed further. We conclude that the HCF of (38220, 196) = 196.
(iii) 1651 and 2032
Answer:
1651 and 2032
When 2032 is divided by 1651, the remainder is 381 to get 2032 = 1651 × 1 + 381
Now consider the division of 1651 with 381 in the above and division algorithm to get 1651 = 381 × 4 + 127
Now consider the division of 381 with the remainder 127 in the above and division algorithm to get 381 = 127 × 3 + 0
The remainder is zero, when we cannot proceed further we conclude that the HCF (1651, 2032) = 127
Question 2.
Use division algorithm to show that any positive odd integer is ofthe form 6q + 1, or 6 q + 3 or 6 q + 5, where q is some integer.
Answer:
Let ‘a’ be any positive odd integer, we apply the division algorithm with a and b = 6. Since 0 ≤ r ≤ 5, the possible remainders are 0, 1, 2, 3, 4 and 5.
i. e., a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where ‘q’ is quotient, However, since ‘a’ is odd, a cannot be 6q, 6q+2, 6q+4 (since they both are divisible by 2)
∴ any odd integer in the form of 6q + 1, (or) 6q + 3 (or) 6q + 5.
Question 3.
Use division algorithm to show that the square of any positive integer is of the form 3p or 3p + 1.
Answer:
Consider ‘a’ be the square of an integer
Applying Euclid’s division lemma with ‘a’ and ‘b’ = 3
0 ≤ r < 3 the possible remainders are 0, 1, 2 a = 3p or 3p + 1 (or) 3p + 2
Any square number is of the form 3p, 3p + 1, or 3p + 2, Where p is the quotient.
Question 4.
Use division algorithm to show that the cube of any positive integer is of the form 9 m, 9m + 1 or 9m + 8.
Answer:
Let ‘a’ be the cube of a positive integer
Applying Euclid’s division lemma for ‘a’ and b = 9 a = bm + r, Where’m’ is quotient r is the remainder where 0 ≤ r < 9 ‘a’ can be of the form
9m, 9m + 1, 9m + 2 ………… or 9m + 8
The cube of any positive integer is of the form 9m, 9m + 1, 9m + 2, ….. or 9m + 8.
Question 5.
Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.
Answer:
Given numbers : n; n + 2 and n + 4
Case (i) : ‘n’ is divisible by 3, then n is of the form 3k. Now, n + 2 = 3k + 2 – leaves a remainder 2 when divided by 3. n + 4 = 3k + 4 = (3k + 3) + 1 = 3(k + 1) + 1 leaves a remainder 1 when divided by 3.
Case (ii) : n + 2 is divisible by 3. then n + 2 is of the form 3k.
Now n = 3k – 2 leaves a remainder 1 when divided by 3 and n + 4 = n + 2 + 2
= 3k + 2 leaves a remainder 2 when divided by 3.
Case (iii) : n + 4 is divisible by 3.
Take n + 4 = 3k
Now : n = 3k – 4 = 3(k – 1) + 3 – 4
= 3(k – 1) – 1 leaves a remainder 2 when divided by 3.
Also n + 2 = 3k-2 = 3(k- 1) + 3-2
= 3(k – 1) + 1 leaves a remainder 1 when divided by 3.
In all the above three cases only one out of n, n + 2 and n + 4 is divisible by 3.