Assuming we have coins with denomination values 1, 5, 10, 25, 50 only. Designate the quantities of each of these respectively as p, n, d, q, h (in US currency this would be penny (cent), nickel, dime, quarter, half-dollar).
p+n+d+q+h=50, the total number of coins. So p=50-(n+d+q+h).
The total value is p+5n+10d+25q+50h=100. So p=100-5(n+2d+5q+10h).
We have two equations for p so we can eliminate p:
100-5(n+2d+5q+10h)=50-(n+d+q+h),
4n+9d+24q+49h=50,
n+9d/4+6q+49h/4=50/4,
n+6q+8d/4+d/4+48h/4+h/4=48/4+2/4,
n+6q+2d+12h+(d+h)/4=12+2/4,
n+(6q+2d+12h-12)=(2-d-h)/4, so 2-d-h must be divisible by 4 and it can be negative.
n=(2-d-h)/4-2(3q+d+6h-6)≥0, because n≥0.
(2-d-h)/4≥2(3q+d+6h-6), 2-d-h≥24q+8d+48h-48,
0≤24q+9d+49h≤50, 9d≤50-49h-24q. So h must be zero and 9d≤50-24q.
If q=0, 0≤d≤5; if q=1, 9d≤26, 0≤d≤2.
(2-d)/4 is an integer only if d=2, so q=0 or 1.
If q=h=0 and d=2, n=8 and p=40; if h=0, q=1 and d=n=2, and p=50-(2+2+1)=45.
So we have two solutions: (p,n,d,q,h)=(40,8,2,0,0) or (45,2,2,1,0).
CHECK: 0.40+0.40+0.20=1.00; 0.45+0.10+0.20+0.25=1.00.