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Operation \(@\) is defined as \(@x = x + 1\) if \(x\) is negative and \(@x = x - 1\) if \(x\) is positive or 0. If \(@(@(@a)) = 5\), what is the value of \(a\) ?

Operation \(@\) is defined as \(@x = x + 1\) if \(x\) is negative and \(@x = x - 1\) if \(x\) is positive or 0. If \(@(@(@a)) = 5\), what is the value of \(a\) ?

Operation \(@\) is defined as \(@x = x + 1\) if \(x\) is negative and \(@x = x - 1\) if \(x\) is positive or 0. If \(@(@(@a)) = 5\), what is the value of \(a\) ?

A. -7 B. -5 C. 2 D. 8 E. 10

Backsolve: \(@(@(@8)) = 8 - 1 - 1 - 1 = 5\).

Answer: D

Hi Bunuel,

@x given and given that if \(@x = x + 1\) if \(x\) is negative and \(@x = x - 1\) if \(x\) is positive or 0.

and given @a , but here how we can consider a as positive.

Since there is no negative sign then you considerd a as +ve ? Please let me know.

Operation \(@\) is defined as \(@x = x + 1\) if \(x\) is negative and \(@x = x - 1\) if \(x\) is positive or 0. If \(@(@(@a)) = 5\), what is the value of \(a\) ?

A. -7 B. -5 C. 2 D. 8 E. 10

Backsolve: \(@(@(@8)) = 8 - 1 - 1 - 1 = 5\).

Answer: D

Hi Bunuel,

@x given and given that if \(@x = x + 1\) if \(x\) is negative and \(@x = x - 1\) if \(x\) is positive or 0.

and given @a , but here how we can consider a as positive.

Since there is no negative sign then you considerd a as +ve ? Please let me know.

The question is solved by backsolving, by substituting the options.
_________________

The symbol represents one step in the number line, whether the numbe is negative or positive. Therefore, since the symbol appears three times, the \(a\) must be either \(a+3\) or \(a-3\). Furthermore, since both hypothetical results are positive (5+3, 5-3), only the former -- 8-- works since all positive numbers are deducted one.

i understood Bunuel's solution. However, i'd a question: how does one decide if one should use back substitution or equations to solve? in this case, i tried solving with equations. since we don't know sign of a, i solved for both and got 2 and 8 as answers then back substituted to check ( since both 2 and 8 are in answer choices) and then found only 9 works. any advise?

i understood Bunuel's solution. However, i'd a question: how does one decide if one should use back substitution or equations to solve? in this case, i tried solving with equations. since we don't know sign of a, i solved for both and got 2 and 8 as answers then back substituted to check ( since both 2 and 8 are in answer choices) and then found only 9 works. any advise?

thanks.

If you take a step back and look at the operation it's pretty simple. If x > 0, subtract 1. If x < 0, add one. The initial number then definitely can't be negative or lower than 5 since you'd then have no way to get up to 5. Hence, if you want to get to 5 you have to start higher, since eventually you'll just keep getting alternating values of 0, -1, 0, -1, 0, -1.

If the operation isn't that simple it's a game time decision I'd think. I typically just backsolve. The only time I'd consider not backsolving is if the answer options are nasty or if the operation is "normal" and just a function - f(k) or f(x) or whatever.

Operation \(@\) is defined as \(@x = x + 1\) if \(x\) is negative and \(@x = x - 1\) if \(x\) is positive or 0. If \(@(@(@a)) = 5\), what is the value of \(a\) ?

A. -7 B. -5 C. 2 D. 8 E. 10

I got the answer right but still link to similar question type please. Thanks.

Operation \(@\) is defined as \(@x = x + 1\) if \(x\) is negative and \(@x = x - 1\) if \(x\) is positive or 0. If \(@(@(@a)) = 5\), what is the value of \(a\) ?

A. -7 B. -5 C. 2 D. 8 E. 10

I got the answer right but still link to similar question type please. Thanks.