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

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 (a^{2}) will be

Case – I: a^{2} – (3q)^{2} = 9q^{2}=3(3q^{2}) = 3p (p = 3q^{2})

Case-II: a^{2} = (3q + l)^{2} = 9q^{2} + 6q+ 1

= 3[3q^{2} + 2q] + 1 = 3p+l (Where p = 3q^{2} + 2q) or

Case – III: a^{2} = (3q + 2)^{2} = 9q^{2} + 12q + 4 = 9q^{2} + 12q + 3 + 1

= 3[3q^{2} + 4q + 1] + 1

= 3p + 1 (where ‘p’ = 3q^{2} + 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.

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 = a^{3} + (9q + r)^{3}

= (9q)^{3} + 3.(9q)^{3}r + 3. 9q.r + r^{3}

= 9^{3}q^{3} + 3.9^{2}(q^{2}r) + 3.9(q.r) + r^{3}

= 9[9^{2}.q^{3} + 3.9.q^{2}r + 3.q.r] + r^{3}

a^{3} = 9m + r^{3} (where ‘m’ = 9^{2}q^{3} + 3.9.q2r + 3.q.r)

if r = 0 ⇒ r^{3} = 0 then a^{3} = 9m + 0 = 9m

and for r = 1 ⇒ r^{3} = l^{3} then a^{3} = 9m + 1

and for r = 2 ⇒ r^{3} = 2^{3} then a^{3} = 9m + 8

for r = 3 ⇒ r^{3}, = 3^{3} ⇒ a^{3} = 9m + 27 = 9(m) where m = (9m +3)

for r = 4 ⇒ r^{3} = 4^{3} ⇒ a^{3} = 9m + 64 = (9m + 63) + 1 = 9m + 1

for r = 5 ⇒ r^{3} = 125 ⇒ a^{3} = 9m + 125 = (9m + 117) + 8 = 9m + 8

for r = 6 ⇒ r^{3} — 216 ⇒ a^{3} = 9m + 216 = 9m + 9(24) = 9m

for r = 7 ⇒ r^{3} = 243

⇒ a^{3} = 9m + 9(27) = 9m

for r = 8 ⇒ r^{3} = 512

⇒ a^{3} = 9m + 9(56) + 8 = 9m + 8

So from the above it is clear that a^{3} is either in the form of 9m or 9m + 1 or 9m + 8.

Hence proved.

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.