What is the smallest number divisible by 5 6 and 7 and 8 leaves remainder 3?
Answers
Answered by
0
LCM of 5, 6, 7 and 8 = 840
If each of these numbers must leave the remainder 3,
add 3 to the LCM ⇒ 840 + 3 = 843
Hence the smallest number divisible by 5, 6, 7 and 8 is 843.
Similar questions