PS: functions - difficult and time consuming one

Please repost - large portions of the question are missing.

Thanks!

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

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.

GMATGuruNY wrote:

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

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.

hi,
is there any other method to solve this problem, instead of directly putting in the values?
pls explain by that method also...

heymayank08 wrote:

GMATGuruNY wrote:

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

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.

hi,
is there any other method to solve this problem, instead of directly putting in the values?
pls explain by that method also...

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.

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

In the given form, how can

2^(x1) * 3^(y1) * 5^(z1)

be 16 times

2^(x2) * 3^(y2) * 5^(z2)?

Since 16 = 2^4, the only way to form "16" in our problem is with 4 "2"s. It follows that x1 = x2 + 4
Since the rest of the terms have to be equal and 3 and 5 are both primes, y1 = y2 and z1 = z2
So the hundreds digit of c needs to be 4 greater than the hundreds digit of k, choose A