Problema Solution
alexandra gives 1/e of her marbles to tyler,who gives 2/3 of what he receives to jan, who gives 3/4 of what she receives to jerry. If each has a counting number of marbles, what is the fewest number of marbles that alexandra could have started with?
Answer provided by our tutors
let 'x' be the number of marbles that Alexandra started with
x(1/e) she gave to Tyler
Tyler gave x(1/e)(2/3) = (2x)/(3e) to Jan
Jan gave ((2x)/(3e))(3/4) = (6x)/(12e) = x/(2e) to Jerry
they all have integer number of marbles thus
x/e is integer
(2x)/(3e) in integer
x/(2e) is integer
the lest common multiple of e, 3e and 2e is 6e thus the fewest number of marbles that Alexandra could have started with is 6*e.