Problema Solution
One box has 15 calculatorsˏ there are 144 batteries in one box. one calculator needs 4 batteries. How many boxes of calculators are needed in order not to have any batteries or calculator left?
Answer provided by our tutors
let
c = the number of boxes with calculators, c is integer
b = the number of boxes with batteries, b is integer
c*15 = 144*b
c*3*5 = 2*2*2*2*3*3*b
Greatest Common Factor of 15 and 14 is 2*2*2*2*3*3*5 = 720
for
c = 720/15 = 48
b = 720/144 = 5
we have
48*15 = 144*5
thus one solution is 48 boxes of calculators and 15 boxes of batteries.
We will prove that the solution is
c = 48k
b = 5k
for every non-negative integer k
indeed
48k*15 = 5k*144
720k = 720k
we got identity thus c = 48k boxes of calculators and b = 5k boxes of batteries, for k positive integer is the solution.