Pascal's n-Cube Triangle

The number of $m$ -cube elements on the boundary of an $n$ -cube is defined as $f(n, m) = 2^{n-m} \binom{n}{m}$ (see inductive proof below). Therefore, the number of squares on the boundary of a $1000$ -cube would be when $m = 2$ and $n = 1000$ , and would be defined as:

$f(1000, 2)$

$= 2^{1000-2} \binom{1000}{2}$

$= 2^{998} \frac{1000!}{(1000 - 2)! \cdot 2!}$

$= 2^{998} \frac{1000!}{998! \cdot 2!}$

$= 2^{998} \frac{1000 \cdot 999}{2}$

$= 2^{998} \frac{10^3 \cdot 999}{2}$

$= 2^{998} \frac{2^3 \cdot 5^3 \cdot 999}{2}$

$= 2^{998 + 3 - 1} \cdot 5^3 \cdot 999$

$= 2^{1000} \cdot 5^3 \cdot 999$

Since $5^3 \cdot 999$ is not divisible by $2$ , the greatest power of $2$ that evenly divides the total number of squares on the boundary of a $1000$ -cube is $2^{1000}$ , and its exponent is $\boxed{1000}$ .

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Inductive Proof:

Assuming $f(n, m) = 2^{n-m} \binom{n}{m}$ is true, it can be shown that $f(n + 1, m) = f(n, m - 1) + 2f(n, m) = 2^{n+1-m} \binom{n+1}{m}$ as follows:

$f(n + 1, m)$

$= f(n, m - 1) + 2f(n, m)$

$= 2^{n-(m - 1)} \binom{n}{m-1} + 2 \cdot 2^{n-m} \binom{n}{m}$

$= 2^{n+1-m} \frac{n!}{(n-(m-1))!(m-1)!} + 2^{n+1-m} \frac{n!}{(n-m)!m!}$

$= 2^{n+1-m}[\frac{n!}{(n+1-m)!(m-1)!} + \frac{n!}{(n-m)!m!}]$

$= 2^{n+1-m}[\frac{n!m}{(n+1-m)!m(m-1)!} + \frac{n!(n+1-m)}{(n+1-m)(n-m)!m!}]$

$= 2^{n+1-m}[\frac{n!m}{(n+1-m)!m!} + \frac{n!(n+1-m)}{(n+1-m)!m!}]$

$= 2^{n+1-m}\frac{n!(m + n + 1 - m)}{(n+1-m)!m!}$

$= 2^{n+1-m}\frac{n!(n + 1)}{(n+1-m)!m!}$

$= 2^{n+1-m}\frac{(n + 1)!}{(n+1-m)!m!}$

$= 2^{n+1-m}\binom{n + 1}{m}$

Therefore, by inductive proof the number of $m$ -cube elements on the boundary of an $n$ -cube is defined as $f(n, m) = 2^{n-m} \binom{n}{m}$ .