Question 1040980
Let N = the possible number of pennies of Joyce in the bank.

==> N = 5r + 1 for some non-negative integer r.
==> N = 3s + 2 for some non-negative integer s.

==> 5r + 1 = 3s + 2 ===> 5r - 3s = 1, 

which is a linear Diophantine equation.

Now any linear Diophantine equation 

ar - bs = c 

has an integer solution in r and s if and only if  
gcd
(a,b) divides c, and that all integer solutions are of the form

{{{r = r[0] - (bk)/gcd(a,b)}}} and {{{s = s[0] + (ak)/gcd(a,b)}}}, where {{{r[0]}}} and {{{s[0]}}} are particular solutions.

gcd(5,3) = 1 divides c = 1 ==> there are general solutions for the DE equation.
A particular solution to this equation is {{{r[0] = 2}}} and {{{s[0] = 3}}}.

==> r = 2 + 3k and s = 3 + 5k,
for k = 0, 1, 2, 3, 4, ...
are the general solutions.

r = 2 and s = 3 (when k = 0) give the lowest possible value for N which is  

N = 11,

and therefore 11 is the least possible number of pennies in the bank.