If z is an integer and z! is divisible by 340, what is the smallest

Question

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

Bunuel · Answer

z!=340*k=2^2*5*17*k

z! must be divisible by ALL the prime factors of 340. If z<17, z! won't have 17 as it's multiple thus won't be divisible by 340. And if z=17 z! is divisible by all the factors of 340.

So, minimum value of z is 17

Hussain15 · Answer

Correct answer. Thanks for explanation.

Posted from my mobile device

GMATD11 · Answer

4) If z is an integer and z! is divisible by 340, what is the smallest possible value for z?

z!/340

z ill minimum have 2*2*5*17

smallest possible value of z =2

But guys OA is different its 17

Pls tell me the logic for the correct answer.

Thanks

Bunuel · Answer

Merging similar topics.

DavidXu · Answer

What if instead of being divisible by 340, it is asking to be divisible by 720?

720 = 2, 2, 2, 2, 2, 3, 3, 5

Thx in advance.

manalq8 · Answer

the same thing, you need to use prime factorization for 720. let's see,

z!= 720k=2.2.2.2.3.3.5k

in this case, depends on what choices you have. see what number in the answer choice has the same prime factors as the ones above for 720. or do little multiplication to get one of the numbers in the answer choices.

hope that helps

kostyan5 · Answer

In that case you simply right out factorials of increasing number until all factors of 720 are also in the factorial.

2*3*4*5*6 = 2*3*(2*2)*5*(2*3) , all 2's, 3's, and 5's are accounted for.

Therefore, smallest z for which z! is divisible by 720 is 6. 6! is 720.

kanusha · Answer

I do not understand red color part

As Prime factors of 720 = 2^4 x 3^2 x 5. From this how can i know factorial?

Thank you

bumpbot · Answer

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If z is an integer and z! is divisible by 340, what is the smallest

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If z is an integer and z! is divisible by 340, what is the smallest

Prep Toolkit

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