We have that
E + B + J < 40
And....when all the trading of counters has ceased..we find that
{E - E/5) = (B - B/3 + E/5 + J/4) = (J - J/4 + B/3)
(4/5)E = [ (2/3)B + E/5 + J/4 ] = [ (3/4)J + B/3 ]
Using the first and third parts of the equality, we have
(4/5)E = B/3 + (3/4)J multiply through by 60→ 48E = 20B + 45J (1)
Using the first two parts of this equality, we have
(4/5)E = [ (2/3)B + E/5 + J/4 ]
(3/5)E = (2/3)B + J/4 multiply through by 60 → 36E = 40B + 15J
Then multiply through by -3 → -108E = -120B - 45J (2)
Add (1) and (2)
-60E = -100B
B = (3/5)E
And using the first two parts of the equality with substitution for B we have that
(3/5)E = (2/3)(3/5)E + J/4
(3/5)E = (2/5)E + J/4
(1/5)E = J/4
J = (4/5)E
So
E + B + J < 40
E + (3/5)E + (4/5)E < 40
(12/5)E < 40
E < 16+2/3
So......since E is divisible by 5, the greatest that Emma could have started with is 15
Which means that Ben started with (3/5)(15) = 9
And Jack started with(4/5)(15) = 12
And
E + B + J =
15 + 9 + 12 =
36 < 40
So....they could have started with 36 counters
BTW.....there are other possibilites
E = 10 , B = 6 and J = 8 = 24 counters
E = 5, B = 3 and J = 4 = 12 counters