Find the highest power of $4$ in $82! + 83!$ I'am only getting $4^{13}$ as answer, but the correct answer is $40$.
What am I missing?
 A: We first find the maximum power of $2$ in $82!+83!=82!(1+83)=82!(84)$, since $83$ is odd we want the largest power of $2$ in $84!$.
Using the Legendre-Polignac formula this is:
$\sum\limits_{n=1}^{\infty}\lfloor\frac{84}{2^n}\rfloor=\lfloor\frac{84}{2}\rfloor+\lfloor\frac{84}{4}\rfloor+\lfloor\frac{84}{8}\rfloor+\lfloor\frac{84}{16}\rfloor+\lfloor\frac{84}{32}\rfloor+\lfloor\frac{84}{64}\rfloor=42+21+10+5+2+1=81$.
So the maximim power of $4$ that divides the number is $4^{40}$
A: $82! = 82\cdot81\cdot80...2!$
41 numbers between 1 and 81 are divisible by 2.
20 are divisible by 4.
10 are divisible by 8.
$41+20+10+5+2+1 = 79$
$2^{79}$ divides $82!$ i.e. $82! = 2^{79} n$, with n odd.
and $2^{79}$ divides $83!, 83! = 2^{79} (83n)$ 
$82! + 83! = 2^{79} (84n) = 2^{81}(21n)$
$4^{40}$ is the largest power of $4$ that divides $2^{81}(21n)$
A: $82! + 83! = 82!(1 + 83) = 82!84 = 82!*4*21$.
$\{1,2,3....82\}$ has:  $41$ multiples of 2.  So $2^41|82!$  Of those $20$ are multiples of 4. so $2^{41+20}|82!$.  Of those $10$ are multiples of 8 so $2^{41+20+10}|82!$.   Of those $5$ are multiples of 16 so $2^{41+20+10+5}|82$.  Of those $2$ are multiples of  of 32    and $1$ is a multiple of 64.
So $2^{41+20+10+5+2+1}=2^{79}|82!$ and $2^{81}|(82! + 83!)$.  So $4^{40}|(82! + 83!)$.  So the answer is $40$.
