/sci/ - Science & Math

>>10023466
is this correct (assuming [math]n+1 \ge r[/math]) ?

[math]a = (n + 1)! + r[/math]

so [math]a = bq + r[/math], where [math]b=r[/math] and [math]q[/math] is all the terms in [math](n+1)![/math] except for [math]r[/math]

and thus [math]gcd((n+1)! + r, b) = gcd(b, r) = gcd(r, r) = r[/math]