Question 1115578
X-number of 5c coins
Y-number of 10c coins
Z-number of 20c coins
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"18 coins"
{{{X+Y+Z=18}}}
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"three times as many 10c coins as 20c coins"
{{{Y=3Z}}}
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{{{X+3Z+Z=18}}}
{{{X+4Z=18}}}
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Since X is divisible by 2 and 3, it is a multiple of 6 and the smallest value it can be is 6.
Since Y and Z are also divisible by 3, the smallest values they can have are 3.
Start with {{{X=6}}},
{{{6+4Z=18}}}
{{{4Z=12}}}
{{{Z=3}}}
So then,
{{{Y=3(3)}}}
{{{Y=9}}}
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(6,9,3) is a solution.
If you try {{{X=12}}},
{{{12+4Z=18}}}
{{{4Z=6}}}
Not an integer solution.
So there is only one solution.