Question

1. What is the greatest number which will divide 110 and 128 leaving a remainder 2 in each case?
2. What is the greatest number that will divide 307 and 330 having remainders 3 and 7 respectively?​

Answer 1

Answer:

Reminder 2. Therefore 110–2=108 AND 128–2=126 are fully divisible.

Answer 2

Step-by-step explanation:

1. 110-2=108

128-2=126

2. 307-3=304

330-7=323

hcf of 304 and 323 is 19.