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25 Jun 2023, 02:00
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Difficulty:
75%
(hard)
Question Stats:
54% (01:49) correct
46%
(01:45)
wrong
based on 459
sessions
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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
A. 0
B. 1
C. 2
D. 3
E. 4
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19 Mar 2024, 07:00
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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
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
General Discussion
25 Jun 2023, 02:14
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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
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
