Number Properties Chapter 1 Question 18

[jzh200]

Question 18: If z is an integer and z! is divisible by 340, what is the smallest possible value for z?

I understand that we need to list the prime factors of 340 (2x5x2x17)

I guess I am getting tripped up on why 17 is the smallest possible value for z? If we look at the prime box (2,5,2,17) wouldn't 2 rather than 17 be the smallest possible value? Perhaps I am misunderstanding the answer, but I do understand part of the explanations (In order for z! to be divisible by 5, z must be atleast 5) yet I can't see why 12 would be the correct answer.

Your input would be appreciated.

[guy29]

Ok think of it this way: z! is supposed to be divisible by 340, which means that z! must be divisible by all the prime factors of 340. Z! in turn must contain 2^2, 5, and 17. Now z is the highest number of z!, if z was say 5, z! would be 5*4*3*2*1, which would not be divisible by 340, because it is missing the prime factor 17; in fact 5! is 120, and you can easily see that it is not divisible by 340. If z where say, 34, then z! would be 34*33*...*1, and while being divisible by 340, it certainly isn't the smallest number z can be. So 17 in turn is the smallest number for z, which for z! would contain every prime factor of 340.

Remember if x is divisible by y, all of the factors of y must be factors of x.

[esledge]

When I try to remember rules like this one correctly presented by guy29...

Remember if x is divisible by y, all of the factors of y must be factors of x.

...I find that it is all too easy for me to get it reversed (something wrong such as "factors of x must be factors of y" or "some factors of x and y must be the same"). If you trust your memory as little as I trust mine, you might benefit from going back to basics: divisibility is simply about division--translate the words into division form.

"z! is divisible by 340" translates to z!/340 = integer, or z!/(2*2*5*17) = integer. You'll only get an integer result from the division if you can cancel ALL of those factors out of the denominator. Therefore, z! must have a factor of 17, and z must be at least 17.