SOLUTION: Jeremy has a bag of peanuts. He is going to share them so he counts them out. First he uses groups of 2, then groups of 3, 5 and 8, but each time there is 1 peanut left. How many a

Algebra ->  Sequences-and-series -> SOLUTION: Jeremy has a bag of peanuts. He is going to share them so he counts them out. First he uses groups of 2, then groups of 3, 5 and 8, but each time there is 1 peanut left. How many a      Log On


   

Question 1030735: Jeremy has a bag of peanuts. He is going to share them so he counts them out. First he uses groups of 2, then groups of 3, 5 and 8, but each time there is 1 peanut left. How many are in the bag? Is there more than one answer?
Please give me some ideas to solve this problem. Thanks !
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
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