prove that : 200!/(10!)^20 × 19!
Answers
Let N=nd=200!(10!)20⋅19!. To show N is an integer, consider the set S of all primes that divide the denominator. For each p∈S, we must show that the highest power of p dividing the numerator is at least as much as the highest power of p dividing the denominator.
Let ep(n) denote the highest power of p dividing n. By de Polignac’s formula De Polignac's formula
ep(n!)=∑k≥1⌊npk⌋. …(⋆)
Note that S={2,3,5,7,11,13,17,19}. If p∈{11,13,17,19}, then ep(d)=1 and ep(n)≥1.
We use eqn. (⋆) to prove that ep(n)≥ep(d) for p∈{2,3,5,7}.
e_2(n) = ⌊2002⌋+⌊2004⌋+⌊2008⌋+⌊20016⌋+⌊20032⌋+⌊20064⌋+⌊200128⌋
=100+50+25+12+6+3+1=197.
e2(d)=20⋅e2(10!)+e2(19!)
=20(⌊102⌋+⌊104⌋+⌊108⌋)+⌊192⌋+⌊194⌋+⌊198⌋+⌊1916⌋
=20(5+2+1)+(9+4+2+1)=176.
We may similarly compute
e3(n)=66+22+7+2=97,e3(d)=20(3+1)+(6+2)=88.
e5(n)=40+8+1=49,e5(d)=(20⋅2)+3=43.
e7(n)=28+4=32,e7(d)=(20⋅1)+2=22.
This completes the verification, and the proof that N is an integer