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.

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

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

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