The least common multiple of $x$, $10$ and $14$ is $70$. What is the greatest possible value of $x$?
First, we prime factorize the given numbers: \[10=2\cdot5, \quad 14=2\cdot7, \quad 70=2\cdot5\cdot7.\]  Since the least common multiple of $10$ and $14$ is already $70$ ($2 \cdot 5 \cdot 7$), we can maximize $x$ by letting it be $\boxed{70}$.
70
Level 4
Prealgebra
Problem: The least common multiple of $x$, $10$ and $14$ is $70$. What is the greatest possible value of $x$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the least common multiple (LCM) of $x$, $10$, and $14$, we first determine the prime factorizations of $10$ and $14$. The prime factorization of $10$ is $2 \times 5$, and for $14$, it is $2 \times 7$. The LCM is found by taking the highest power of each prime that appears in these factorizations.

The primes involved are $2$, $5$, and $7$. The highest powers are:
- For $2$: $2^1$ (from both $10$ and $14$)
- For $5$: $5^1$ (from $10$)
- For $7$: $7^1$ (from $14$)

Thus, the LCM of $10$ and $14$ is calculated as:
LCM($10$, $14$) = $2^1 \times 5^1 \times 7^1 = 70$.

Since the LCM of $x$, $10$, and $14$ is also $70$, $x$ must be a divisor of $70$. The divisors of $70$ are $1$, $2$, $5$, $7$, $10$, $14$, $35$, and $70$. 

To maximize $x$, we select the largest divisor, which is $70$. We must verify that the LCM of $70$, $10$, and $14$ remains $70$. 

Calculating the LCM of $70$, $10$, and $14$:
- The prime factorization of $70$ is $2^1 \times 5^1 \times 7^1$.
- The LCM will still be $2^1 \times 5^1 \times 7^1 = 70$.

Since $70$ satisfies the condition that the LCM of $x$, $10$, and $14$ equals $70$, the greatest possible value of $x$ is confirmed to be $70$. 

The final answer is $70$.<|END_OF_LATENT|> The final answer is $70$.