Four consecutive positive integers have a product of 840. What is the largest of the four integers?
We have $840=2^3\cdot3\cdot5\cdot7$. From this prime factorization, it is clear that the product of four consecutive positive integers is $840=2^2\cdot5\cdot(2\cdot3)\cdot7=4\cdot5\cdot6\cdot7$. The largest of the four integers is $\boxed{7}$.
7
Level 2
Number Theory
Problem: Four consecutive positive integers have a product of 840. What is the largest of the four integers?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find four consecutive positive integers whose product is 840, we denote the integers as n, n+1, n+2, and n+3. The product can be expressed as n(n+1)(n+2)(n+3) = 840. 

First, we estimate the value of n by taking the fourth root of 840, since the product of four numbers close to n should be around n^4. The fourth root of 840 is approximately 5.5, suggesting that n could be around 5 or 6.

Next, we test n = 5:
5 * 6 * 7 * 8 = 1680, which is too high.

Now, we test n = 4:
4 * 5 * 6 * 7 = 840, which is correct.

The integers are 4, 5, 6, and 7. Among these, the largest integer is 7. 

Thus, the final answer is 7.<|END_OF_LATENT|> The final answer is $7$.