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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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