AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.1

10th Class Maths 1st Lesson Real Numbers Ex 1.1 Textbook Questions and Answers

Question 1.
Use Euclid’s division algorithm to find the HCF of
i) 900 and 270
Answer:
900 = 270 × 3 + 90
270 = 90 × 3 + 0
∴ HCF = 90

ii) 196 and 38220
Answer:
38220 = 196 × 195 + 0
∴ 196 is the HCF of 196 and 38220.

iii) 1651 and 2032
Answer:
2032 = 1651 × 1 + 381
1651 = 381 × 4 + 127
381 = 127 × 3 + 0
∴ HCF = 127

AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1

Question 2.
Use Euclid division lemma to show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integers.
Answer:
Let ‘a’ be an odd positive integer.
Let us now apply division algorithm with a and b = 6.
∵ 0 ≤ r < 6, 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 the quotient.
But ‘a’ is taken as an odd number.
∴ a can’t be 6q or 6q + 2 or 6q + 4.
∴ Any odd integer is of the form 6q + 1, 6q + 3 or 6q + 5.

Question 3.
Use Euclid’s division lemma to show that the square of any positive integer is of the form 3p, 3p + 1.
Answer:
Let ‘a’ be the square of an integer.
Applying Euclid’s division lemma with a and b = 3
Since 0 ≤ r < 3, the possible remainders are 0, 1, and 2.
∴ a = 3q (or) 3q + 1 (or) 3q + 2
∴ Any square number is of the form 3q, 3q + 1 or 3q + 2, where q is the quotient.
(or)
Let ‘a’ be a positive integer
So it can be expressed as a = bq + r (from Euclideans lemma)
now consider b = 3 then possible values of ‘r’ are ‘0’ or ‘1’ or 2.
then a = 3q + 0 = 3q (or) 3q + 1 or 3q + 2 now square of given positive integer (a2) will be
Case – I: a2 – (3q)2 = 9q2=3(3q2) = 3p (p = 3q2)
Case-II: a2 = (3q + l)2 = 9q2 + 6q+ 1
= 3[3q2 + 2q] + 1 = 3p+l (Where p = 3q2 + 2q) or
Case – III: a2 = (3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1
= 3[3q2 + 4q + 1] + 1
= 3p + 1 (where ‘p’ = 3q2 + 4q + 1)
So from above cases 1, 2, 3 it is clear that square of a positive integer (a) is of the form 3p or 3p + 1
Hence proved.

AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1

Question 4.
Use Euclid’s division lemma to show that the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8.
(OR)
Show that the cube of any positive integer is of form 9m or 9m + 1 or 9m + 8, where m is an integer.
Answer:
Let ‘a’ be positive integer. Then from Euclidean lemma a = bq + r;
now consider b = 9 then 0 ≤ r < 9, it means remainder will be 0, or 1, 2, 3, 4, 5, 6, 7, or 8
So a = bq + r
⇒ a = 9q + r (for b = 9)
now cube of a = a3 + (9q + r)3
= (9q)3 + 3.(9q)3r + 3. 9q.r + r3
= 93q3 + 3.92(q2r) + 3.9(q.r) + r3
= 9[92.q3 + 3.9.q2r + 3.q.r] + r3
a3 = 9m + r3 (where ‘m’ = 92q3 + 3.9.q2r + 3.q.r)
if r = 0 ⇒ r3 = 0 then a3 = 9m + 0 = 9m
and for r = 1 ⇒ r3 = l3 then a3 = 9m + 1
and for r = 2 ⇒ r3 = 23 then a3 = 9m + 8
for r = 3 ⇒ r3, = 33 ⇒ a3 = 9m + 27 = 9(m) where m = (9m +3)
for r = 4 ⇒ r3 = 43 ⇒ a3 = 9m + 64 = (9m + 63) + 1 = 9m + 1
for r = 5 ⇒ r3 = 125 ⇒ a3 = 9m + 125 = (9m + 117) + 8 = 9m + 8
for r = 6 ⇒ r3 — 216 ⇒ a3 = 9m + 216 = 9m + 9(24) = 9m
for r = 7 ⇒ r3 = 243
⇒ a3 = 9m + 9(27) = 9m
for r = 8 ⇒ r3 = 512
⇒ a3 = 9m + 9(56) + 8 = 9m + 8
So from the above it is clear that a3 is either in the form of 9m or 9m + 1 or 9m + 8.
Hence proved.

AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1

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.
(Or)
Show that one and only one out of a, a + 2 and a + 4 is divisible by 3 where ‘a’ is any positive integer.
Answer:
Let ‘n’ be any positive integer.
Then from Euclidean’s lemma n = bq + r (now consider b = 3)
⇒ n = 3q + r (here 0 ≤ r < 3) which means the possible values of ‘r’ = 0 or 1 or 2
Now consider r = 0 then ‘n’ = 3q (divisible by 3)
and n + 2 = 3q + 2 (not divisible by 3)
n + 4 = 3q + 4 (not divisible by 3)
Case – II: For r = 1
n = 3q + 1 (not divisible by 3)
n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + l) divisible by 3
n + 4 = 3q + 1 + 4 = 3q + 5 not divisible by 3
Case – III: For r = 2,
n = 3q + 2 not divisible by 3
n + 2 = 3q + 2 + 2 = 3q + 4, not divisible by 3
n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) divisible by 3
So in all above three cases we observe, only one of either (n) or (n + 1) or (n + 4) is divisible by 3.
Hence proved.

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