Answer:
There can be a minimum of 211 balls present in the bucket.
Explanation:
Given the marbles can be counted by twos, threes, five and seven
Therefore, taking lcm of 2, 3, 5, 7
LCM = [tex]2\times 3\times 5\times 7 = 210[/tex]
So, 210 balls can be counted in 2’s, 3’s, 5’s and 7’s and it is given that exactly one ball is left after counting
Hence, fewest no. of balls that can be present in the basket = 210+1 = 211
Therefore, there can be a minimum of 211 balls present in the bucket.
