ORIGINAL QUESTION.
Let N be the number of screws for 80c, so 1 screw costs 80/N.
After reduction N+1 screws cost 80-3=77c, so 1 screw costs 77/(N+1).
77/(N+1)÷80/N=77N/(80(N+1)) is the reduction as a fraction. 77/80=0.9625 so the reduction is 0.9625N/(N+1)=0.9625/(1+(1/N)). This corresponds to a reduction of over 3.75%, which is greater than 1%. There must therefore be a greater reduction than 1%. (If N=the minimum of 1, the reduction is 1-0.9625/2=51.875%. If N is very large the reduction approaches the minimum of 1-0.9625=0.0375=3.75% from a maximum of 51.875%.)
For example, if the reduction was 5.5%, 0.945=0.9625/(1+1/N); 0.945+0.945/N=0.9625; 0.945=0.0175N, N=54.
Original cost of screw=80/54=40/27=1.4815c.
REVISED QUESTION
Let C=cost of screw, and N be the number of screws, NC=80.
When the cost of a screw is C-1, then (N+1)(C-1)=80-3=77.
So NC-N+C-1=77; 80-N+C=78; N-C=2. Therefore N=C+2. C(C+2)=80; C^2+2C-80=0; (C+10)(C-8)=0 so C=8 and N=10. The original price of one screw is 8c.