If x is an integer greater than 1, is y divisible by x?

Question

(1) y is divisible by x − 1
(2) y = x! + x − 1

Hatakekakashi · Answer

statement 1 - y mod x-1 = 0

y mod x can be =0 or != 0

i.e y=100 x=5 x-1 = 4 the answer is yes
 y = 27 x=10 x-1 = 9 the answer is no

so CBD

statement 1 is not sufficient

statement 2 - y = x! + x - 1

we know y mod x = 0 + 0 - 1

y= kx - 1 where k is any integer

so statement 2 is sufficient

(B) imo

rahulkashyap · Answer

if y=4 and x=2 and x-1 = 1, ans is yes
if y=5 and x=2 and x-1 = 1 , ans is no

x! and x are both divisble by x, but 1 is not divisible by x, as x>1

ans should be b

jch103020 · Answer

Here is my approach:

Stem: x is integer >1, is y/x an integer?

1) y is divisible by x-1 ---> y/(x-1) = integer

Insufficient: x = 4 --> y/3 = integer
 x = 3 --> y/2 = integer
 Both of these cases can't be true, thus (1) is insufficient

2) y=x!+x-1

x! = x(x-1)

y=x(x-1)+x-1
y=x^2-x+x-1
y=x^2-1
y=x(x-1)
y/(x-1)=x

Sufficient
----

Can anyone comment on my reasoning for deciding statement 1 is insufficient? I got the correct, answer, but want to be sure I didn't just get lucky with my reason for eliminating A

globaldesi · Answer

This problem rewards students who remember that all multiples of the integer x
 will be x
 units apart. Statement (1) gives that y
 is divisible by x−1
. This can be proven insufficient with a quick number picking thought experiment. If x=4
 is it possible to have a number that is divisible by both x=4
 and x−1=3
? If you multiply 3 and 4 together to get 12, the answer is yes, there are an infinite number of ways to have a value of y
 that is divisible by both x
 and x−1
.

Conversely, there are also an infinite number of integers that are divisible by x−1
 that aren't divisible by x
. In the case of x−1=3
, 6, 9, and 21 are all divisible by 3 but aren't divisible by 4. Since this means you can get both a "yes" and a "no" based on the information given in statement (1), statement (1) is insufficient. Eliminate (A) and (D).

Statement (2) gives that y=x!+x−1
. This may seem to give information similar to what is given in statement (1), but it is much more sufficient.

Remember that x!
 can be written as "(x)(x−1)(x−2)(x−3).....(2)(1)
".

This means that for any given value for x
 where x
 is an integer greater than 1, that x!
 must be divisible by x−1
 as well as x
.

If you add x−1
 to x!
, you get a number that is divisible by x−1
 but that cannot be divisible by x
. To understand this, it helps to pick numbers.

If x=4
, then x!=(4)(3)(2)=24
. If you add x−1=3
 to that, you get 27, which is still divisible by 3 but that isn't divisible by 4.

This is because multiples of the same number are evenly spaced - all multiples of 4 will be 4 apart, all multiples of 3 will be 3 apart, etc. So for whatever number you pick for x
, if you add x−1
 you will always get a number that is 1 less than a multiple of x
, which means that y
 can never be divisible by x
.

This means that statement (2) gives you a consistent "no" and is sufficient. Answer choice (B) is correct.

Mo2men · Answer

the question asks if  \frac{y}{x}  =Integer

(1) y is divisible by x − 1

Let x =2...then x-1 = 1........ \frac{y}{1}  =  y

If y odd..........answer is NO

If y even......... answer is Yes

Insufficient

(2) y = x! + x − 1

Divide both sides by x

\frac{y}{x} = \frac{x!}{x} + \frac{x}{x} − \frac{1}{x}

It is clear that every term will yield integer value except  \frac{1}{x} .......Answer is always NO

Sufficient

Another approach to plug-in values

Let x =2 .........y=3............Answer is NO

Let x= 3..........y=8...........Answer is NO

Let x = 4.........y=27..........Answer is NO

Let x= 5..........y =124........Answer is No

Answer: B

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If x is an integer greater than 1, is y divisible by x?

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