Dint even understand the question!

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The function g is defined as follows. For any 3 digit integer (written xyz), Function (xyz) is 2^x.3^y.5^z
. If c and k are 3 digit integers, and function (C) is 16 (f)K what is the value of c - k?

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Hi shibsriz,

This question can be beaten by TESTing values and using the information that your'e given to limit the possibilities.

We have 3 variables (x, y and z) that form a 3-digit number: xyz

**REMINDER: 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 (which we can see in the function).

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 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

[GMATGuruNY]

The function f is defined as follows:

For any 3 digit integer (written xyz), f(xyz)=2^x3^y5^z.

If c and k are 3 digit integers, and f(c)=16*f(k), what is the value of c - k?

(A) 400

(B) 320

(C) k/16c

(D) 40

(E) Cannot be determined

Approach 1:

For every 3-digit integer xyz, f(xyz) = (2^x)(3^y)(5^z).

Let f(k) = 100.
Thus:
f(xyz) = 100.

Substituting f(xyz) = (2^x)(3^y)(5^z) and prime-factorizing the righthand side, we get:
(2^x)(3^y)(5^z) = 2²3�5².

Since the bases on each side of the equation match, so must the corresponding exponents.
Thus, x=2, y=0, z=2.
Thus, k = (3-digit integer xyz) = 202.

Since f(c) = 16f(k) and f(k)=100, f(c) = 16(100).
Thus:
f(xyz) = 16(100).

Substituting f(xyz) = (2^x)(3^y)(5^z) and prime-factorizing the righthand side, we get:
(2^x)(3^y)(5^z) = (2�)(2²3�5²) = 2�3�5².

Since the bases on each side of the equation match, so must the corresponding exponents.
Thus, x=6, y=0, z=2.
Thus, c = (3-digit integer xyz) = 602.

Thus, c-k = 602-202 = 400.

The correct answer is A.

Approach 2:

In 3-digit integer k, let H = the hundreds digit, T = the tens digit, and U = the units digit.
Thus, 3-digit integer k = HTU = 100H + 10T + U.

f(k) = f(HTU) = (2^H)(3^T)(5^U).

f(c) = 16*f(k) = 2^4 * (2^H)(3^T)(5^U) = 2^(4+H)*(3^T)*(5^U).
Since the exponents here represent the 3 digits of c:
3-digit integer c = [4+H]TU = 100(4+H) + 10T + U = 400 + 100H + 10T + U.

c - k = (400 + 100H + 10T + U) - (100H + 10T + U) = 400.

[Matt@VeritasPrep]

One addendum: while the original prompt is hard to follow, if I'm reading it correctly (and I may not be) there are a lot of possible values for c and k, though I agree with Rich that (c - k) = 400.

Let's define the function first.

If xyz is a three digit number such that x is the 100s digit, y is the 10s digit, and z is the units digit (e.g. xyz is NOT x * y * z), then g(xyz) = 2^x * 3^y * 5^z.

So, for instance, g(209) = 2^2 * 3^0 * 5^9.

If we know that c and k are three digit integers, let c = the three digit # abd and k = the three digit # efg.

f(c) = f(abd) = 2^a * 3^b * 5^d
f(k) = f(efg) = 2^e * 3^f * 5^g

If f(c) = 16f(k), then 2^a * 3^b * 5^d = 16 * 2^e * 3^f * 5^g,
or
2^a * 3^b * 5^d = 2^4 * 2^e * 3^f * 5^g
or
2^a * 3^b * 5^d = 2^(e+4) * 3^f * 5^g

Since both sides of the equation represent integers, and these integers are the same, they have the same prime factorization. Our bases are all prime, so a = e + 4, b = f, and d = g.

Returning to our three digit numbers, that means that they each have the same tens and units digits, but the hundreds digit of c is 4 greater than the hundreds digit of k. So c - k = 400.

Just to illustrate, say c = 512 and k = 112. Then f(c) = 2^5 * 3^1 * 5^2 and f(k) = 2^1 * 3^1 * 5^2. f(c) = 16f(k), and c and k are three digit integers, so these values satisfy the equation.

Open follow up question to stimulate a chat on these boards: how many possible coordinates (c,k) are there that satisfy the function above and the equation f(c) = 16f(k)?

EDIT: Mitch, you are quick on the draw, you beat me to it! Still, I took forever typing this, so I'm leaving it up