## Topic outline

• ### Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.7)

• Playing With Numbers

Exercise 2.7 (Continued Division Method – Euclid’s Algorithm)

 Question 1 (Refer Book - Mathematics Class VI R.D. Sharma) Solution:

Clearly, the last divisor is 150.

Hence the H.C.F of the given number is 150.

Clearly, the last divisor is 19.

Hence the H.C.F of the given number is 19.

Clearly, the last divisor is 95.

Hence the H.C.F of the given number is 95.

 Question 2 (Refer Book - Mathematics Class VI R.D. Sharma) Solution: i. 59, 97 Factors of 59 - 1 and 59 Factors of 97 - 1 and 97 Common Factor = 1 Therefore, H.C.F is 1. So, 59 and 97 are co-prime.   ii. 875, 1859 Factors of 875 – 1, 5, 7, 25, 35, 125, 175, 875 Factors of 1859 – 1, 1859 Common Factor = 1 Therefore, H.C.F is 1. So, 875 and 1859 are co-prime.   iii. 288, 1375 Factors of 288 – 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 48, 72, 96, 144, 288. Factors of 1375 – 1,5,11,25,55,125,275,1375 Common Factor = 1 Therefore, H.C.F is 1. So, 288 and 1375 are co-prime. Question 3 (Refer Book - Mathematics Class VI R.D. Sharma) Solution: H.C.F of two consecutive numbers is 1. Example: 3,4 Factors of 3 – 1, 3 Factors of 4 – 1, 2, 4 Common Factor – 1 Therefore, H.C.F = 1 Question 4 (Refer Book - Mathematics Class VI R.D. Sharma) Solution: i. True ii. False iii. True iv. True v. False