76. Is the number of combinations of 1000 elements taken 500 at a time divisible by 7?
76. The number of combinations of 1000 elements taken 500 at a time is $\frac{1000!}{(500!)^{2}}$: Since 7 is a prime number, the highest power of 7 that divides 1000! (see the second solution to problem 756) is $\left[\frac{1000}{7}\right]+\left[\frac{1000}{49}\right]+ \left[\frac{1000}{343}\right]=142+20+2=164$. The highest power of 7 that divides 500! is $\left[\frac{500}{7}\right]+\left[\frac{500}{49}\right]+\left[\frac{500}{343}\right]=71+10+1=82$. Therefore, the highest power of 7 that divides the denominator is $82 \cdot 2=164$. Thus, both the numerator and the denominator contain 7 to the 164th power. After simplification, the numerator will no longer contain 7 as a factor, and therefore, the integer $\frac{1000!}{(500!)^{2}}$ is not divisible by 7.