Question 394013
The probability that at most one of n coins show tails is 

{{{nC0*(1/2)^n*(1/2)^0 + nC1*(1/2)^(n-1)*(1/2)^1}}}
= {{{(1/2)^n + n(1/2)^n}}}
= {{{(n+1)(1/2)^n}}}.

Now the above sequence STRICTLY DECREASES as n INCREASES, because

{{{(n+1)/2^n < n/2^(n-1) }}} for {{{n > 1}}}.

Values for n =1,2,3,4,5 will not satisfy  {{{(n+1)(1/2)^n = 3/16}}}, but n = 6 will.  Since the sequence {{{(n+1)(1/2)^n}}} is strictly decreasing, 6 is the only value of n that will satisfy the problem.

Therefore there were 6 coins.