If x is an integer, how many solutions exist for the equation |x| + |x

Question

If x is an integer, how many solutions exist for the equation  |x| + |x + 1| = |x|!  ?

A. 0
B. 1
C. 2
D. 3
E. 4

Bunuel · Accepted Answer

Official Solution:

If  x  is an integer, how many solutions exist for the equation  |x| + |x + 1| = |x|! ?

A. 0
B. 1
C. 2
D. 3
E. 4

The equation may appear daunting at first glance, yet, a closer examination reveals a simpler path to solution. Notice that the left-hand side invariably results in an odd number (when  x  is odd,  x + 1  becomes even, and their sum is odd; similarly, if  x  is even, then  x + 1  is odd, resulting in an odd sum). On the other side, the right-hand side represents the factorial of a non-negative integer, which is only odd for the cases of  0! = 1  and  1! = 1 . For any other number, the factorial becomes even, as it includes a factor of 2. Consequently, we only need to consider those values of  x  for which  |x|! = odd = 1 , specifically, -1, 0, and 1.

Substituting these values into the equation shows that  x=0  and  x=-1  are the only valid roots. Thus, the given equation has two solutions.

Answer: C

mysterymanrog · Answer

First thing to note: Do not try to solve using algebra.

Now, notice that the solutions are 0-4. This means not many integers, so we can plug and test values. Notice that the factorial will increase more than the sum of two consecutive integers after a certain value -> 4+5=9, but 4!=24. Way bigger. So the values of X will be quite small 
Notice also that no restrictions have been placed on X -> x may be negative or positive (there are no negative factorials, however since we have modulus x factorial, x may be negative)

Lets test values:
x=0 -> 0+1=0! 
0!=1 by construction, so this is 1 solution. 
x=1 -> 1+2=1! , no the sum is larger 
x=2 -> 2+3=2!, no the sum is larger 
x=3 -> 3+4=3!, no the sum larger 
x=4 -> 4+5=4!, no the factorial is larger, and for any values of x bigger it will never be equal because of how quickly factorials grow.

Don't forget negatives: 
x=-1 -> 1+0=1!, yes 2 solutions 
x=-2 -> 2+1=2!, no the sum is larger 
x=-3 -> 3+2=3!, no the sum is larger 
x=-4 -> 4+3=4!, no the factorial is larger -> this is the end like in the positive case

We have a total of two solutions (x=-1, x=0) therefore the correct answer choice is C

Bunuel · Answer

Great work. However, there's a more efficient method that requires testing only -1, 0, and 1. Can you identify why this is the case?

gmatophobia · Answer

The LHS of the equation |x| + |x+1| will always be an odd integer.

Reason  : x and x + 1 are consecutive integers. Hence, the sum of their distances from 0 will always be odd.

The RHS of the equation is a factorial.

a! is odd only when the value of a = 1 or 0. In all other cases (i.e. when a  \geq  2), the value of the factorial is even.

As, a = |x|, x can be either 0, 1, or -1.

We can use the above values to check which of the values satisfies the equation.

x = 0

|x| + |x + 1| = |x|!

|0| + |0 + 1| = |0|!

0 + 1 = 1

Works!  Keep x = 0

x = 1

|1| + |1 + 1| = |1|!

|1| + |2|  
eq  |1|!

Discard x = 1

x = -1

|-1| + |-1 + 1| = |-1|!

1 + 0 = 1!

Works!  Keep x = -1

Possible values of x = {-1,0}

Option C

TheBipedalHorse · Answer

|x| + |x+1| = |x|!

By observation- the right hand side |x|! is always even unless x = 0, or x = -1 or x = +1

The left hand is always odd (adding two consecutive numbers)

Testing solutions for -1, 0 and +1

At x = -1, LHS is 1, RHS is 1.....<bingo>
At x = 0, LHS is 1, RHS (0!) is 1.....<bingo>
At x = 1, LHS is 3, RHS is 1.....<no>

Therefore, 2 solutions

JOBS07 · Answer

Both X=0 & X=-1 create mirror images in the LHS, and, on the RHS, the values anyhow are same. If x=1, the equality doesn't hold. Anything beyond these extreme points, the LHS is a sum of consecutive numbers (always odd) and the RHS is always even (as the lowest possible value is 2, which is even), a situation wherein LHS can never be equal to RHS.

Bunuel · Answer

Check similar question here: https://gmatclub.com/forum/if-x-is-an-i ... 14316.html

bumpbot · Answer

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If x is an integer, how many solutions exist for the equation |x| + |x

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If x is an integer, how many solutions exist for the equation |x| + |x

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