<PRE><B>Question 1030735</B>
2, then groups of 3, 5 and 8, but each time there is 1 peanut left.

The number of peanuts must be 1 more than the multiples of all the above 

2*3*5*8  = 120  

121 peanuts is one possibility

You can form groups if you have 1 more than any multiple of 120 which is the LCM






 
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