find the smallest number which when divided by 8,9,10,15,20 gives a remainder of every time
Answers
Answered by
2
Answer:
6 is the answer I think it is right
Answered by
3
Answer:
The required number is 361
Explanation:
The smallest number exactly divisible by 8, 9, 10, 15 and 20
= L C M of 8, 9, 10, 15 and 20.
= 360
Therefore the required number
= 360 + 1
= 361 Ans.
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