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# For positive integers x and y, x=19!+y. Is x a prime number?

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Math Expert
Joined: 02 Sep 2009
Posts: 59722
For positive integers x and y, x=19!+y. Is x a prime number?  [#permalink]

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18 Apr 2017, 05:12
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Difficulty:

55% (hard)

Question Stats:

58% (01:28) correct 42% (01:21) wrong based on 97 sessions

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For positive integers x and y, x=19!+y. Is x a prime number?

(1) 41>y>37
(2) y is even

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Re: For positive integers x and y, x=19!+y. Is x a prime number?  [#permalink]

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18 Apr 2017, 05:32
2
Given data -> x=19!+y
We need to see if x is prime or not.

Statment 1=>
y can be 38,39,40

If y=38 => x=2*(some integer) => Non prime
y=39=> x=13*(some integer) => Non prime
y=40 => x=2*(some integer) => Non prime

Basically the logic used here is that PRIMES cannot be subdivided (other than off course--> Prime*1).
Hence x is not a prime.

Hence Sufficient.

Statement 2=>
If y is even => x=2*(some integer)=> Again => Non prime.

Hence x can never be prime.
Hence Sufficient.

Smash that D.

NOTE-> all some integers are greater than 1

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Re: For positive integers x and y, x=19!+y. Is x a prime number?  [#permalink]

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18 Apr 2017, 05:42
Bunuel wrote:
For positive integers x and y, x=19!+y. Is x a prime number?

(1) 41>y>37
(2) y is even

Hi..

$$x=19!+y$$...
When will x be prime Or even better and easier here when will it be NOT a prime
when y contains any factor SAME as of 19! And 19! Has all factors from 2 to 19

Let's see the statements..
1) 41>y>37..
Y can be 38,39 or 40..
If 38 or 40, Common factor will be 2..
If 39, common factor will be 3..
In every case and is NO
Sufficient

2) y is Even..
In every case, common factor will be 2..
So x is multiple of 2
Ans is NO always
Sufficient

D
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Re: For positive integers x and y, x=19!+y. Is x a prime number?  [#permalink]

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24 Nov 2019, 11:07
Bunuel wrote:
For positive integers x and y, x=19!+y. Is x a prime number?

(1) 41>y>37
(2) y is even

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 2 variables and 1 equation, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
The possible values of $$y$$ are 38, 39, 40.
If $$y=38=2*19$$, we have $$19!+38 = 19! + 2*19 = 18!*19 + 2*19 = (18!+2)*19$$, which is not a prime number.
If $$y=39=3*13$$, we have $$19!+39 = 1*2*3*4*...*19 + 3*13 = 3*(2*4*5*...*19) + 3*13 = 3*((2*4*5*...*19)+13)$$, which is not a prime number.
If $$y=40=2*20$$, we have $$19!+39 = 1*2*3*4*...*19 + 2*20 = 2*(3*4*5*...*19) + 2*20 = 2*((3*4*5*...*19)+20)$$, which is not a prime number.

Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient.

Condition 2)
If $$y=2*k$$, we have $$19!+2k = 1*2*3*4*...*19 + 2k = 2*(3*4*5*...*19) + 2k = 2*((3*4*5*...*19)+k)$$, which is not a prime number.

Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, condition 2) is sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: For positive integers x and y, x=19!+y. Is x a prime number?   [#permalink] 24 Nov 2019, 11:07
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