minimum value of sum of two number for which binomial coefficients is an even integer

Question

for two non negative integer values of $r$ say $r_{1},r_{2}(r_{1}\neq r_{2})$ for which $\displaystyle \binom{1999}{r}$ is an even integer

Then the minimum value of $r_{1}+r_{2}$

$\displaystyle \binom{1999}{r}$ represent coefficient of $x^r$ in $(1+x)^{1999}$

i wan,t be able to go further, could some help me with this, Thanks

Answer 1

For any prime number $p$ and non-zero interger $n$, the $p$-adic valuation of $n$, denoted by $v_p(n)$, is the exponent of $p$ in the prime factorization of $n$. Notice

• $P1$ - $p$ divides $n$ if and only if $v_p(n) > 0$.
• $P2$ - $v_p(nm) = v_p(n)v_p(m)$.
• $P3$ - for $0 < n < p^m$, $v_p(n) = v_p(p^m N \pm n)$ for any integer $N$ and $m > 0$.

By $P1$, the problem at hand is equivalent to finding the first two $r$ such that $$v_2\left(\binom{1999}{r}\right) = v_2\left(\frac{1999!}{r!(1999-r)!}\right) = v_2\left(\prod_{k=1}^r\frac{2000-k}{k}\right) > 0\tag{*1}$$

By $P2$, for any non-zero integers $a_1, \ldots, a_k$, $b_1, \ldots, b_k$. If $N = \frac{a_1 \ldots a_A}{b_1 \ldots b_B}$ is an integer, we have $$v_2(N) = \sum_{i=1}^A v_2(a_i) - \sum_{j=1}^B v_2(b_j)$$

Using this, the condition in $(*1)$ is equivalent to

$$\sum_{k=1}^r (v_2(2000-k) - v_2(k)) > 0$$

Notice $2000 = 2^4\cdot 125$. We can use $P3$ to conclude

$$v_2(2000-k) - v_p(k) = 0\quad\text{ for } 1 \le k \le 15$$

This implies $r \ge 16$.

• When $k = 16$, we have $v_2(2000-k) - v_2(k) = v_2(2^6\times 31) - v_2(2^4) = 6 - 4 = 2$
• When $k = 17$, $2000-k$ and $k$ are both odd, so $v_2(2000-k) - v_2(k) = 0$.

Combine these, we get

$$v_2\left(\binom{1999}{r}\right) = \begin{cases} 0, & 1 \le r \le 15,\\ 2, & r = 16, 17 \end{cases}$$

The $r_1, r_2$ we seek are at least $16$ and $17$ and hence the minimum value of $r_1 + r_2$ is $16+17 = 33$.

Notes

• For a proof of $P3$, write $n$ as $p^e f$ where $e = v_p(n)$. By definition of $v_p(n)$, $\gcd(p,f) = 1$.
  Rewrite $p^m N \pm n$ as $p^e(p^{m-e} N \pm f)$. Since $$p^e \le n < p^m \implies e < m \implies p | p^{m-e} N \implies \gcd(p^{m-e} \pm f, p ) = \gcd(f,p) = 1$$ $e = v_p(n)$ is also the exponent for $p$ in the prime factorization of $p^m N \pm n$.