Part A: Other Divisors of $X$

If 10 and 25 are divisors of $X$, then their greatest common divisor (GCD) and least common multiple (LCM) are important in determining the other divisors.

The prime factorizations are:

• $10 = 2 \times 5$
• $25 = 5^2$

For a number to be divisible by both 10 and 25, $X$ must contain the prime factors of both numbers. Thus, $X$ must include:

• At least one factor of 2 (from 10),
• At least two factors of 5 (from 25).

From this, we know that $X$ must include at least the factors:

• $5$ (common in both 10 and 25),
• $2$ (from 10),
• $25$ (from 25).

Thus, $X$ must also have the divisors:

• 1 (all integers are divisible by 1),
• 2 (since 2 is a factor of 10),
• 5 (since 5 is a factor of both 10 and 25),
• 50 (since $50 = 2 \times 25$).

Part B: Three Possible Values for $X$

$X$ must be divisible by the least common multiple (LCM) of 10 and 25. The LCM of 10 and 25 is found as:

• $LCM(10, 25) = 2 \times 5^2 = 50$.

Thus, $X$ must be at least 50. Some possible values for $X$ (multiples of 50) include:

• $X = 50$,
• $X = 100$,
• $X = 150$.

Part C: The Smallest Value $X$ Could Be

The smallest value $X$ could be is the least common multiple of 10 and 25, which we calculated to be: $X = 50$

Summary:

1. Divisors of $X$: $1, 2, 5, 10, 25, 50$
2. Three possible values for $X$: 50, 100, 150
3. Smallest possible value for $X$: 50

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Do you want more details or have any further questions?

Related Questions:

1. What is the process for finding the least common multiple of two numbers?
2. How does prime factorization help in finding common divisors?
3. Why is the greatest common divisor (GCD) useful in problems like this?
4. What other methods can you use to find all divisors of a number?
5. How do divisibility rules help in solving such problems?

Tip: To find divisors quickly, perform prime factorization and then consider all combinations of the factors.