The value of $\sqrt{73}$ is between two positive, consecutive integers. What is the product of these two integers?
Since $73$ is between $64=8^{2}$ and $81=9^{2}$, we know that $\sqrt{73}$ is between $8$ and $9.$ Our answer is $8\cdot9=\boxed{72}.$
72
Level 2
Prealgebra
Problem: The value of $\sqrt{73}$ is between two positive, consecutive integers. What is the product of these two integers?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the two consecutive integers between which $\sqrt{73}$ lies, we first identify the perfect squares around 73. The perfect square of 8 is $8^2 = 64$, and the perfect square of 9 is $9^2 = 81$. Since $64 < 73 < 81$, we conclude that $\sqrt{73}$ is between 8 and 9.

Next, we identify the two consecutive integers as 8 and 9. The product of these two integers is calculated as follows: 

8 multiplied by 9 equals 72. 

Thus, the final answer is 72.<|END_OF_LATENT|> The final answer is $72$.