integers

saadiagha wrote:If Z is an integer and Z! is divisible by 340, what is the smallest possible value of z?

Hey guys, was stuck on this question, I thought I got the answer correct but it didint seem to match the answer given in the book.

Preamble: A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7

So, for Z! to be divisible by 340, it must be the case that 340 is hiding in the prime factorization of Z!

Since 340 = (2)(2)(5)(17), we can see that the prime factorization of Z! must include a 17 (and two 2's and a 5).
This tells us that the smallest possible value of Z is 17.

Cheers,
Brent

saadiagha wrote:hey Brent

thanks for the answer- I got till the part where I took out the prime factors and concluded that z has the same factors as 340. However, I dont understand how 17 is the least possible value of z.

Shoudnt 2 be the least possible value? Considering its the smallest number.

No problem.

Let's start with Z = 2
So, z! = (1)(2)
We're missing a 2, a 5 and a 17 in the prime factorization of Z!

How about Z = 3
So, z! = (1)(2)(3)
We're still missing a 2, a 5 and a 17 in the prime factorization of Z!

Z = 4
So, z! = (1)(2)(3)(4) = (1)(2)(3)(2)(2)
We're missing a 5 and a 17 in the prime factorization of Z!

Z = 5
So, z! = (1)(2)(3)(4)(5) = (1)(2)(3)(2)(2)(5)
We're still missing a 17 in the prime factorization of Z!

Z = 6
So, z! = (1)(2)(3)(4)(5)(6) = (1)(2)(3)(2)(2)(5)(2)(3)
We're still missing a 17 in the prime factorization of Z!
.
.
.

Since 17 is a prime number, we won't get a 17 in the prime factorization of Z! until Z = 17

Cheers,
Brent