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Divisibility and primes

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by ramannjit » Fri Oct 01, 2010 2:52 am
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]
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by GMATGuruNY » Fri Oct 01, 2010 3:01 am
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.
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by ramannjit » Fri Oct 01, 2010 3:18 am
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?
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by GMATGuruNY » Fri Oct 01, 2010 3:23 am
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
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For more information, please email me (Mitch Hunt) at [email protected].
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by ramannjit » Fri Oct 01, 2010 3:26 am
Great! crystal clear.

Thanks a lot!:)
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