let f(x) be defined as the integer k such that

Question

Let f(x) be defined as the integer k such that 10^k \le x < 10^{k+1} . If f(y) = 4 and z > y , which of the following must be true? ⸻ I. f\left(\frac{z+y}{2}\right) \le \frac{1}{2}\big(f(z) + f(y)\big) II. f(yz) < f(y)f(z) III. f\big(f(z^y)\big) > 4 ⸻ (A) I only (B) II only (C) III only (D) I and II (E) II and III

Devikaudcaps · Answer

can someone post a solution to this

is this a high quality question or can ignore

GMATcracker101 · Answer

Please provide solution. I am unable to figure out how to solve this. Thank you.

kevincan · Answer

When can we say about the value of y ?

GMATcracker101 · Answer

As per the question, what I can understand is the following - Since f(y) = 4 as per the equation in the question, 10^4 ≤y<10^5 w.k.t z>y that implies that the integer that defines f(z) must be greater than the integer that defines f(y), i.e., 4. Hence f(z) > 4 Using this to solve I. II. and III. gives - I. f(z+y/2) ≤ 1/2*(f(z) + 4) II. f(yz) < 4*f(z) III. f(f(z^y)) > 4 ---- can't really say anything about this one. I am not sure how to use this to solve I. II. and III. Please provide any further hints, thanks.

kevincan · Answer

So we know that y is a four-digit positive integer and that z > y

GMATcracker101 · Answer

Right, yes. So we can also say that in II. f(yz) 10^8 ≤ yz < 10^9 => f(yz) = 8 (minimum) and f(z) is at least 4. So, II stands true.

kevincan · Answer

As for I, there’s no limit as to high z is . Imagine if it were a 10 digit integer

GMATcracker101 · Answer

Understood now, thank you.

Adit_ · Answer

The statement 1 here is False. Why?
Because z can be any value and there is no limit. The function value if you take both as 4 digit numbers will provide you with answer as 4. But for z being 10 digit number can be much higher thus is it not lesser than or equal to RHS in this case.

Statement 2 on the other hand is True. Why?
Because product of any large number multiplied by a 5 digit number may increase the no.of digits of the max value by 6 digits at max. But simply multiplying the function together leads to the k value ins 10^k and LHS is thus definitely going to be smaller here.

Statement 3: This is again false. I believe you can work this out yourself using principles of number of digits.
 Well one can see that by taking the smallest 5 digit number in this case 10001^10000.
Here can you spot that 4 digits after 0 will be added each time the exponent moves from 1 to 10000
Meaning the number of digits is definitely going to be atleast 4*10000 = 40,000 more longer which is an insane number. Lets for satisfaction case even take this number to be 50,000 digits.
What is f(50,000) now? It is simply 4. As k value is 4 in this case. Function will give us 5 only if the value of x inside the function exceed 10^5. Thus the whole LHS value is 4 and it is NOT greater than 4. Thus FALSE.

_________________________________

This is honestly pretty hard considering the function looks intimidating. Take your time with it.

makennaben · Answer

how should one go about answering this question if seeing something like this for the first time

Kinshook · Answer

Let f(x) be defined as the integer k such that 10^k \le x < 10^{k+1} . If f(y) = 4 and z > y , which of the following must be true? 10^4=10000 \le y < 10^5=100000 . z > y > 10000 f(z) > 4 ⸻ I. Let z = 900000; f(z) = 5 f\left(\frac{z+y}{2}\right) \le \frac{1}{2}\big(f(z) + f(y)\big) f (\frac{10000+900000}{2}) = f(455000) = 5 \frac{1}{2} (4+5) = 4.5 5 is NOT < 4.5 NOT NECESSARILY TRUE II. Let f(z) = m > 4 10^m < z < 10^{m+1} f(yz) < f(y)f(z) 10^{m+4} < yz < 10^{5+m+1} = 10^{m+6} Maximum value of f(yz) = m+5 f(y) = 4; f(z) = m; f(y)f(z) = 4m m+5 < 4m m > 5/3 Since m>4> 5/3 MUST BE TRUE III. f\big(f(z^y)\big) > 4 Let z = 10001 & y = 10000 z^y = (10001)^{10000} f(z^y) = 40000 f(f(z^y) = 4 is NOT > 4 NOT NECESSARILY TRUE (A) I only (B) II only (C) III only (D) I and II (E) II and III IMO B

egmat · Answer

Hi makennaben,

The key with any scary-looking custom function is to not react to the notation. Two moves handle almost every one of these, and they're exactly what kevincan was nudging the thread toward.

Step 1 - Translate the definition into plain words

f(x) = the k where 10^k ≤ x < 10^(k+1). That's just asking: "which power-of-10 band does x live in?" For a positive integer, that's simply (number of digits − 1).

So the givens become concrete:
- f(y) = 4 → 10^4 ≤ y < 10^5
- z > y → z > 10^4 → f(z) ≥ 4

Step 2 - For "must be true," try to BREAK each statement

"Must be true" means true in every case. So your real job is to hunt for one counterexample. Find one and the statement is dead. The best hunting ground is extreme values - try z barely above y, then z gigantic.

I: let y = 10^4, z = 10^10. Then (z+y)/2 ≈ 5×10^9, so f = 9, but the right side is (10+4)/2 = 7. Is 9 ≤ 7? No - I dies. (This is exactly kevincan's "imagine z is a 10-digit number.")

III: let y = 10^4, z = 2×10^4. Then f(z^y) ≈ 43010, and f(43010) = 4, not greater than 4 - III dies.

II: you can't break this. Multiplying a band-4 number by z lands in a band of at most f(z) + 5, while the right side is 4·f(z). Since f(z) ≥ 4, we get f(z) + 5 < 4·f(z) every time - II survives.

Lock in the habit

Decode a fresh custom function fast, in plain words: "g(x) = the number of times you can halve x before dropping below 1." Immediately test it - g(8) = ?, g(9) = ? - before touching the statements.

Translate first, then attack must-be-true with extreme test numbers. That combo cracks the whole family.

Answer: B

makennaben

how should one go about answering this question if seeing something like this for the first time

let f(x) be defined as the integer k such that

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