So she has 5N+1 pennies, where N is some positive integer

So she has 3M+2 pennies, where M is some other positive integer.

So 3M+2 = 5N+1 
   3M-5M = -1

Write 5 in terms of its nearest multiple of 3, which is 6

   3M-(6-1)N = -1
     3M-6N+N = -1

Divide through by 3

      M-2N+N/3 = -1/3

Get fractions left, whole numbers right:

       N/3+1/3 = 2N-M

Right side is a positive integer, so the right side is too.
Let that positive integer by A

N/3+1/3 = A;      2N-M = A

   N+1 = 3A;      
     N = 3A-1;   

Substitute in 2N-M = A

2(3A-1)-M = A
   6A-2-M = A
       5A = M+2
     5A-2 = M

So N = 3A-1 and M = 5A-2

The least possible value of A to make N the
least positive integer is A=1.

So N = 3A-1 = 3(1)-1 = 3-1 = 2 and M = 5A-2 = 5(1)-2 = 5-2 = 3

So she has 5N+1 pennies = 5(2)+1 = 10+1 = 11 pennies

As a check:

She also has 3M+2 pennies = 3(3)+2 = 11 pennies.

And when we divide 11 by 5, we get 1 remainder, and
when we divide 11 by 3, we get 2 remainder.

So 11 pennies is correct.

Edwin