find the smallest number which when divided by 35,56 and 91 leaves remainders of 7 in each case
Answers
Answered by
3
Answer:
Step-by-step explanation:
The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647. 2) The number which is exactly divisible by 8, 15 and 21 = LCM of 8, 15 and 21.
Similar questions