Inequality Qn

[anantbhatia]

If x is an integer, does x have a factor n such that 1 < n < x?

(1) x > 3!

(2) 15! + 2 ≤ x ≤ 15! + 15


[spoiler]
OA B. Grockit[/spoiler]

[neerajkumar1_1]

statement 1)
x > 3!
3!= 3* 2*1 ..... 3,2,1 are factors
4!= 4*3*2*1 ..... 4,3,2,1 are factors

so x will always have a factor between 1<n<x
sufficient

Statement 2)
15!+2<=x<=15!+15

now 15! + 2 = 15*14*13....*2*1 + 2 = 2(15*14*13.... +1)
so 2 is a factor of 15!+2
u can do the same for 3,4,5.... 15

hence there will always be a factor
sufficient

IMO: D

[abhi.genx7]

Statement 1 says x is greater than 3! but does not mention that the number must be a factorial , so it can be a prime too like 11 .
I thought in this way .
And was too lazy to solve the 2nd one .
Can i have the OA please?

[narik11]

If x is an integer, does x have a factor n such that 1 < n < x?

(1) x > 3!

(2) 15! + 2 ≤ x ≤ 15! + 15


[spoiler]
OA B. Grockit[/spoiler]

Option 1 is not enough as x can be 7

Option 2 is enough as 15!+2 is nothing but multiple of 2.
15! has 2 as one of the factor. so when (2) is added to 15! it is divisible by 2.
Likewise for all the numbers till 15.

[anantbhatia]

Thanks narik. You made it look so simple. And it was so..

Can anyone please tell me a generic method to solve inequalities. I am bogged down by them and then resort to plugging.