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If Z is an integer, is Z prime?
(1) \(15! < Z\)
(2) \(17! + 2 \leq Z \leq 17! + 17\)
(1) \(15! < Z\)
(2) \(17! + 2 \leq Z \leq 17! + 17\)
If Z is an integer, is Z prime?
(1) \(15!<z\) --> \(z\) is more than some number (\(15!\)). \(z\) may or may not be a prime. Not sufficient.
(2) \(17!+2\leq{z}\leq{17!+17}\) --> \(z\) cannot be a prime. For instance if \(z=17!+13=13*(2*3*4*5*6*7*8*9*10*11*12*14*15*16*17+1)\), then \(z\) is a multiple of 13, so not a prime. Same for all other numbers in this range. So, \(z=17!+x\), where \(2\leq{x}\leq{17}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(17!+x\), the same way as we did for 13). Sufficient.
Answer: B.
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Hope it's clear.
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(1) \(15!<z\) --> \(z\) is more than some number (\(15!\)). \(z\) may or may not be a prime. Not sufficient.
(2) \(17!+2\leq{z}\leq{17!+17}\) --> \(z\) cannot be a prime. For instance if \(z=17!+13=13*(2*3*4*5*6*7*8*9*10*11*12*14*15*16*17+1)\), then \(z\) is a multiple of 13, so not a prime. Same for all other numbers in this range. So, \(z=17!+x\), where \(2\leq{x}\leq{17}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(17!+x\), the same way as we did for 13). Sufficient.
Answer: B.
Similar questions to practice:
if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html
does-the-integer-k-have-a-factor-p-such-that-1-p-k-126735.html
Hope it's clear.
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@ akhileshgupta05:
Satment 1 is not sufficient
For statement 2 possible values of z are
17! + 2 is Divisible by 2 . Hence not prime
17! + 3 is Divisible by 3 . Hence not prime
17! + 4 is Divisible by 4 . Hence not prime
17! + 5 is Divisible by 5 . Hence not prime
17! + 6 is Divisible by 6 . Hence not prime
17! + 7 is Divisible by 7 . Hence not prime
17! + 8 is Divisible by 8 . Hence not prime
17! + 9 is Divisible by 9 . Hence not prime
17! + 10 is Divisible by 10 . Hence not prime
17! + 11 is Divisible by 11 . Hence not prime
17! + 12 is Divisible by 12 . Hence not prime
17! + 13 is Divisible by 13 . Hence not prime
17! + 14 is Divisible by 14 . Hence not prime
17! + 15 is Divisible by 15 . Hence not prime
17! + 16 is Divisible by 16 . Hence not prime
17! + 17 is Divisible by 17 . Hence not prime
In any case z cannot not be prime hence answer is B
Satment 1 is not sufficient
For statement 2 possible values of z are
17! + 2 is Divisible by 2 . Hence not prime
17! + 3 is Divisible by 3 . Hence not prime
17! + 4 is Divisible by 4 . Hence not prime
17! + 5 is Divisible by 5 . Hence not prime
17! + 6 is Divisible by 6 . Hence not prime
17! + 7 is Divisible by 7 . Hence not prime
17! + 8 is Divisible by 8 . Hence not prime
17! + 9 is Divisible by 9 . Hence not prime
17! + 10 is Divisible by 10 . Hence not prime
17! + 11 is Divisible by 11 . Hence not prime
17! + 12 is Divisible by 12 . Hence not prime
17! + 13 is Divisible by 13 . Hence not prime
17! + 14 is Divisible by 14 . Hence not prime
17! + 15 is Divisible by 15 . Hence not prime
17! + 16 is Divisible by 16 . Hence not prime
17! + 17 is Divisible by 17 . Hence not prime
In any case z cannot not be prime hence answer is B
General Discussion
subhashghosh
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The answer is B:
17! + 2 <= Z <= 17! + 17 means that z is like 17! + 2, 17! + 3 etc. So you can factor out a common integer from the expressions that will divide 17! + n (2 <= n <=17), e.g. 2 divides 17! + 2 because 17! also includes 2.
So z is not a prime.
(1) is not sufficient because there are primes and non-primes above 15!
Please ask if you have any more queries.
17! + 2 <= Z <= 17! + 17 means that z is like 17! + 2, 17! + 3 etc. So you can factor out a common integer from the expressions that will divide 17! + n (2 <= n <=17), e.g. 2 divides 17! + 2 because 17! also includes 2.
So z is not a prime.
(1) is not sufficient because there are primes and non-primes above 15!
Please ask if you have any more queries.
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If the statement B was rephrased to ===> 17! + 2 < Z < 17! + 19 then will the answer to this question change to E ?
Erjan_S
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I read the Kaplan explanation and read explanations above in the thread and I can't find a clue....
For instance, why 17!+13 is equal to 13∗(2∗4∗5∗6∗7∗8∗9∗10∗11∗12∗14∗15∗16∗17+1)? Isn't (2∗4∗5∗6∗7∗8∗9∗10∗11∗12∗14∗15∗16∗17+1) = 17!? So, why 17! multiplied by 13 would be equal to 17! + 13?
For instance, why 17!+13 is equal to 13∗(2∗4∗5∗6∗7∗8∗9∗10∗11∗12∗14∗15∗16∗17+1)? Isn't (2∗4∗5∗6∗7∗8∗9∗10∗11∗12∗14∗15∗16∗17+1) = 17!? So, why 17! multiplied by 13 would be equal to 17! + 13?
Erjan_S
17! + 13 = 2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17 + 13
Factor out 13: 13*(2*3*4*5*6*7*8*9*10*11*12*14*15*16*17 + 1)
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Bunuel
Bunuel can you please give similar example (with factoring out number) where Z is prime number ?
dave13
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Epex
Bunuel, what is the answer to this?
If statement B was rephrased to 17! + 2 < Z < 17! + 19, wouldn't the answer remain the same? Z < 17! +19, meaning Z can be at most 17! + 18 which is not prime.
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