<PRE><B>Question 416972</B>
I just tried every single number that could be the one left out
First I let {{{x}}} = the amount one person paid and {{{2x}}} = 
amount other person paid. {{{x + 2x = 3x}}}, so the sum of 
what they bought must be divisible by {{{3}}}
The only way I could arrive at this is if the coin costing 20
is left out. Then {{{15 + 16 + 18 + 19 + 31 = 99}}}, so if
{{{x + 2x = 99}}}
{{{3x = 99}}}
{{{x = 33}}}
1 person bought 33 worth of coins
The other bought 66 worth
{{{15 + 18 = 33}}}
and
{{{ 16 + 19 + 31 = 66 }}}
So, he kept the coin costing 20</PRE>