The function f is defined as follows: for any 3-digit integer (written

c-k = (x+4) * 100 + y*10 + z - ( x*100+ y*10 + z)

leaves 400. A.

400

as only powers of 2 can make it 16 times

so hundredth digit must be 4 more

C = pqr

f(c) = 2^p3^5^r

K = lmn

f(k) = 2^l3^m5^n


2^p3^5^r = 16 * 2^l3^m5^n

=> p = l+4, m = 3, n = r

So pqr - lmn = (l+4)100 + 10m + n - (l)100 + 10m + n = 400

Answer - A

Hi All,

This question can be beaten by TESTing VALUES and using the information that you're given to limit the possibilities.

We have 3 variables (X, Y and Z) that form a 3-digit number: XYZ

**NOTE: when we see XYZ that DOES NOT mean multiply X, Y and Z**

The function we're given is this:
f(XYZ) = (2^X)(3^Y)(5^Z)
So, if we have the three digits, then we just just plug them into the function and get a value.

Next, we're told that C and K are both 3-digit numbers and that f(C) = 16(f(K). Now, THAT is interesting because for a number to be 16 times another number, we're going to have to deal with powers of 2 (16 = 2^4). Notice how the math involved uses 2^X......

I'm going to keep things as simple as possible:

K = 100 This is the smallest 3 digit number that is available and it will make our math easy.

f(K) = (2^1)(3^0)(5^0) = (2)(1)(1) = 2

Now we need a result that is 16 times that.... so we have to take the first digit (the 'power' of 2) and raise it, while keeping everything else the same....

C = 500

f(C) = (2^5)(3^0)(5^0) = (32)(1)(1) = 32

Now we have two 3-digit numbers that fit what we're told (and it turns out that they're the ONLY numbers that would fit the given scenario)

C - K =400

Final Answer:

GMAT assassins aren't born, they're made,
Rich

f(c)=16∗f(k) implies that f(c)/f(k) =16.

this can be true if the powers 3 and 5 for f(c) and f(k) are equal and there is a difference of 4 in the powers of 2.

For eg. (2^6 * 3^2 * 5^3) / (2^2 3^2 * 5^3) = 16. Therefore, this gives the numbers as 623 and 223. Difference is 400.

Similarly, we can replace the other permissible values and check. The answer will always be 400.

Cheers!
Sidhant

Functions and custom characters look complicated though they are often not so.

\(f(xyz)=2^x3^y5^z\)

E.g, \(f(432) = 2^4*3^3*5^2 = f(k)\) (Assume values to see what it looks like)

\(f(c)=16*f(k) = 2^4*f(k) = 2^4 * 2^4*3^3*5^2 = 2^8*3^3*5^2\)
So if k = 432, c = 832

c - k = 400

Answer (A)

Can you pls explain the highlighted part?

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