The function f is defined for all positive integers n > 4 as f(n) = 3n - 9 if n is odd and f(n) = 2n - 7 if n is even. What is the value of the positive integer a?

(1) f(f(a)) = a

(2) f(f(f(a))) is odd.

Tough question. The first thing I'll do is pick some numbers to see if I can establish any kind of pattern with this function. I'm told n > 4, so let's start with n = 5. Because 5 is odd, we'll use f(n) = 3n - 9.

f(5) = 3*5 - 9 = 6

Okay, let's try 6. Because 6 is even, we'll use f(n) = 2n - 7.

f(6) = 2*6 - 7 = 5

I'll do a few more:

f(7) = 3*7 - 9 = 12

f(8) = 2*8 - 7 = 9

f(9) = 3*9 - 9 = 18

f(10) = 2*10 - 7 = 13

So I note two things: first, that we alternate between evens and odds. When n is odd, f(n) is even, and vice versa. Second, that f(n) and n are very close when n = 5 or n = 6, but seem to drift further apart as n increases.

(1) f(f(a) = a

Okay. Say a = 5. We'll solve for f(f(5). We've established that f(5) = 6. So if we substitute 6 for f(5), we'll have f(f(5) = f(6). And f(6) = 5. So f(f(5) = 5. 5 is one possible value for a.

Say a =6. We'll solve for f(f(6). f(6) = 5. So if we substitute 5 for f(6), we'll have f(5). We know f(5) = 6. So f(f(6) = 6. 6 is another possible value for a.

Because a could be 5 or 6, this statement alone is insufficient.

(2) (2) f(f(f(a))) is odd.

Well, I know that when a is odd, f(a) is even, and vice versa. So if f(f(f(a))) is odd, then f(f(a) is even. And if f(f(a) is even, then f(a) is odd. And if f(a) is odd, then a is even. Alone this clearly isn't sufficient, because 'a' could be any even number greater than 4.

Together: Now I have two pieces. S1 told me that a = 5 or a = 6. (Any value of 'a' larger than 6 won't satisfy the statement.) S2 told me that a is even. Together, I know that a must be 6. Together they are sufficient. Answer is C.

Hi ,

Thank you so much for the explanation but honestly speaking , still I don't get it.

Please help me out.

Thanks

Thank you so much for the explanation but honestly speaking , still I don't get it.

Please help me out.

Let's break this down into smaller pieces. I offered an approach that involved picking numbers. Ian offered an approach that involved algebra/number properties. Both are valid strategies when grappling with function questions.

First, let's make sure we understand the function.

When n is odd, f(n) = 3n - 9, and when n is even, f(n) = 2n - 7.

If you were picking numbers, you'd start with 5, as we're told the function is true for n > 4.

If n = 5, we're dealing with an odd, so f(n) = 3n - 9, or f(5) = 3*5 - 9 = 6.
So we know f(5) = 6.

If n = 6, we're dealing with an even, so f(n) = 2n - 7, or f(6) = 2*6 - 7 = 5.
So we know f(6) = 5.

If n = 7, we're back to odd, so f(7) = 3*7 - 9 = 12.
So we know f(7) = 12.
etc.

Put another way, we can understand functions in terms of inputs and outputs. If the input is 5, the output is 6. (When n = 5, f(5) = 6.)
Similarly, if the input is 6, the output is 5. (When n = 6, f(6) = 5.)

So I noted two things: first that when my input was ODD, my output was EVEN, and vice versa. And secondly, that when my inputs were smaller, they were close to the outputs.


Ian initially thought about the function in terms of number properties. If n = ODD, then f(n) = 3n - 9. Or f(ODD) = 3*ODD - 9.
3*ODD - 9 = ODD - ODD = EVEN

If n = EVEN, then f(n) = 2n - 7. or f(EVEN) = 2*EVEN - 7 = EVEN - ODD = ODD.

Again, we see that if the input is ODD, the output is EVEN and vice versa.

This is what we want to establish before evaluating the statements. Does that make sense so far? (If it does, I'll go ahead and evaluate the statements in a second post. If not, let's clarify any confusion first.)

j_shreyans wrote:The function f is defined for all positive integers n > 4 as f(n) = 3n - 9 if n is odd and f(n) = 2n - 7 if n is even. What is the value of the positive integer a?

(1) f(f(a)) = a

(2) f(f(f(a))) is odd.

Statement 1: f(f(a)) = a
Case 1: a is odd
If a is odd, then a must be plugged into f(n) = 3n - 9, implying that f(a) = 3a-9.
Thus:
f(f(a)) = f(3a-9).

Since 3a-9 = (3)(odd) - odd = odd - odd = even, 3a-9 must be plugged into f(n) = 2n - 7, implying that f(3a-9) = 2(3a-9) - 7 = 6a - 25.

Since f(f(a)) = 6a - 25 and f(f(a)) = a, we get:
6a - 25 = a
5a = 25
a = 5.

Case 2: a is even
If a is even, then a must be plugged into f(n) = 2n - 7, implying that f(a) = 2a-7.
Thus:
f(f(a)) = f(2a-7).

Since 2a-7 = 2(even) - odd = even - odd = odd, 2a-7 must be plugged into f(n) = 3n - 9, implying that f(2a-7) = 3(2a-7) - 9 = 6a - 30.

Since f(f(a)) = 6a - 30 and f(f(a)) = a, we get:
6a - 30 = a
5a = 30
a = 6.

Since it's possible that a=5 or that a=6, INSUFFICIENT.

Statement 2: f(f(f(a))) is odd.
In Case 1:
a=odd
f(a) = 3a-9, which is even.
f(f(a)) = 5, which is odd.
Notice that the results alternate between odd and even.
Implication:
In Case, 1, since f(f(a)) = odd, f(f(fa))) = EVEN.
Thus, Case 1 does not satisfy statement 2.

In Case 2:
a=even
f(a) = 2a-7, which is odd.
f(f(a)) = 6, which is even.
Notice that the results alternate between even and odd.
Implication:
In Case, 2, since f(f(a)) = even, f(f(fa))) = ODD.
Thus, Case 2 satisfies statement 2.

Case 2 implies that f(f(fa))) = ODD if a is even.
Since a can be any even integer greater than 4, INSUFFICIENT.

Statements combined:
Since only Case 2 satisfies both statements, a = 6.
SUFFICIENT.

The correct answer is C.