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by saadiagha » Tue Dec 18, 2012 1:13 pm
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.
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by Brent@GMATPrepNow » Tue Dec 18, 2012 1:40 pm
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
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by saadiagha » Tue Dec 18, 2012 1:47 pm
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.
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by Brent@GMATPrepNow » Tue Dec 18, 2012 1:52 pm
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
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by saadiagha » Tue Dec 18, 2012 1:58 pm
excellent!

That explains it! Thanks a lot
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by puneetkhurana2000 » Tue Dec 18, 2012 6:50 pm
Answer 17.
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by sharoonsaleem » Fri Dec 21, 2012 5:18 am
This is a good question. Take LCM of 340, it gives out 2,2,5,17.

Now for 340 to divide Z! - Z has to have the highest prime factor. Note that the small factors 2 and 5 shall be taken care of automatically due to the factorial going up till 1 :)

Hence 17 has to be answer.
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