ramannjit wrote:GMATGuruNY wrote:ramannjit wrote:If z is an integer and z! is divisible by 340, what is the smallest possible value for z?
Help me with the above with detailed explaination please.
OA [spoiler]17[/spoiler]
340 = 2*2*5*17
So z! must be divisible by 4, 5 and 17.
Smallest possible factorial that includes 4, 5 and 17 is 17!.
So smallest possible z = 17.
Hi thanks for the explaination. Help me understand further please! what I am unable to understand is if we look at the prime box (2,5,2,17) would not 2 rather than 17 be the
smallest possible value?
If z=2, then z! = 2! = 2*1 = 2. 2 is not divisible by 340.
If z=17, then z! = 17! = 17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1.
The preceding list of factors includes 4, 5 and 17.
Any factorial smaller than 17! will not include 17 among its factors.
So smallest possible z = 17.
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