n=int(input())
d=int(input())
e=int(input())*5
q=n//e
ans=n
for i in range(q+1):
h=(n-i*e)%d
ans=min(ans,h)
print(ans)
#include <cstdio>
typedef long long ll;
int main(){
ll n; scanf("%lld", &n);
ll d, h; scanf("%lld %lld", &d, &h);
h *= 5;
ll mn(n);
for(ll p = 0; p * h <= n; p++){
ll rem = (n - p * h) % d;
mn = (mn < rem) ? mn : rem;
}
printf("%lld\n", mn);
return 0;
}
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