MHB Solving 7x=210: Must x be a Multiple of 12?

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To determine if x must be a multiple of 12 when 7x is a multiple of 210, the prime factorization of 210 is analyzed as 2, 3, 5, and 7. Since 7x = 210n, dividing by 7 gives x = 2 * 3 * 5 * n. This simplifies to x = 30n, indicating that x is a multiple of 30. Therefore, x does not have to be a multiple of 12, as 30 is not a multiple of 12.
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Please help me understand how to get the solution to the following:

If 7x is a multiple of 210, must x be a multiple of 12?

Thank you!
 
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I would begin by finding the prime factorization of 210, which is:

$$210=2\cdot3\cdot5\cdot7$$

Now, if $7x$ is a multiple of 210, then we may write:

$$7x=210n=2\cdot3\cdot5\cdot7\cdot n$$

Now dividing through by 7, we find:

$$x=2\cdot3\cdot5\cdot n$$

What can you conclude from this?
 
I really appreciate your response but I'm getting lost at the "$$7x=210n=2\cdot3\cdot5\cdot7\cdot n$$" mark. Sorry. :(
MarkFL said:
I would begin by finding the prime factorization of 210, which is:

$$210=2\cdot3\cdot5\cdot7$$

Now, if $7x$ is a multiple of 210, then we may write:

$$7x=210n=2\cdot3\cdot5\cdot7\cdot n$$

Now dividing through by 7, we find:

$$x=2\cdot3\cdot5\cdot n$$

What can you conclude from this?
 
If $n$ is a natural or counting number, then the multiples of 210 can be written as $210n$. All we did is then turn the statement "7x is a multiple of 210" into an equation. :D
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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