SEBA Class 10 Maths Solution: Exercise 1.1 (Real Numbers) | SEBA অনুশীলনী ১.১ সমাধা

(i) 135 and 225

Since 225 > 135, we apply Euclid's division lemma:

225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0

The remainder is 0. The divisor at this stage is 45.

Answer: HCF = 45

(ii) 196 and 38220

38220 = 196 × 195 + 0

Answer: HCF = 196

(iii) 867 and 255

867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0

Answer: HCF = 51

Let a be any positive odd integer and b = 6. Possible remainders are 0, 1, 2, 3, 4, 5.

So a can be 6q, 6q+1, 6q+2, 6q+3, 6q+4, 6q+5.

However, 6q, 6q+2, and 6q+4 are even (divisible by 2). Since a is odd, it must be in the form:

6q + 1, 6q + 3, or 6q + 5.

We need to find HCF(616, 32).

616 = 32 × 19 + 8
32 = 8 × 4 + 0

Answer: Maximum 8 columns.

Let integer be x and b = 3. x can be 3q, 3q+1, 3q+2.

• If x = 3q → x² = 9q² = 3(3q²) = 3m
• If x = 3q+1 → x² = (3q+1)² = 9q² + 6q + 1 = 3(3q²+2q) + 1 = 3m + 1
• If x = 3q+2 → x² = (3q+2)² = 9q² + 12q + 4 = 3(3q²+4q+1) + 1 = 3m + 1

Find HCF of 64 and 80.

80 = 64 × 1 + 16
64 = 16 × 4 + 0

Answer: 16 cm

(i) 135 আৰু 225

225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0

উত্তৰ: গ.সা.উ. = 45

(iii) 867 আৰু 255

867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0

উত্তৰ: গ.সা.উ. = 51

ধৰা হ'ল a যিকোনো এটা যোগাত্মক অযুগ্ম অখণ্ড সংখ্যা আৰু b = 6।

ভাগশেষ r = 0, 1, 2, 3, 4, 5 হ'ব পাৰে।

কিন্তু 6q, 6q+2, 6q+4 হ'ল যুগ্ম সংখ্যা। যিহেতু a অযুগ্ম, গতিকে ই কেৱল 6q + 1, 6q + 3 বা 6q + 5 আৰ্হিৰ হ'ব।

616 = 32 × 19 + 8
32 = 8 × 4 + 0

উত্তৰ: 8 টা স্তম্ভ।

64 আৰু 80 ৰ গ.সা.উ. উলিয়াব লাগিব।

80 = 64 × 1 + 16
64 = 16 × 4 + 0

উত্তৰ: 16 ছে.মি.