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The function f is defined as follows: for any 3-digit integer (written 16 May 2011, 19:14
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65% (hard)

Question Stats:

58% (02:13) correct 42% (02:20) wrong based on 431 sessions

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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


©Grockit
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pike
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The function f is defined as follows: for any 3-digit integer (written 16 May 2011, 20:10
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\(c = efg\)
\(k = lmn\)

\(f(c) = 2^e*3^f*5^g\)
\(f(k) = 2^l*3^m*5^n\)

\(f(c) = 16 * f(k)\)
\(2^e*3^f*5^g = 2^4 * 2^l*3^m*5^n\)

So, \(e = (4 + l), f = m, g = n\)

\(c - k = efg - lmn = (4+l)mn - lmn = 400\)

e.g k = 100, f(k) = 2, f(c) = 16*f(k) = 32 = 2^5, therefore c = 500, 500 - 100 = 400
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Spidy001
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The function f is defined as follows: for any 3-digit integer (written 16 May 2011, 20:26
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my solution is pretty similar to what @pike mentioned above

given f(xyz) = \(2^x 3^y 5^z\)

lets say c = efg and k = hij

=> f(c)= \(2^e 3^f 5^g\)
=>f(k) = \(2^h 3^i 5^j\)

given f(c)= 16*f(k)

equating powers we have

e = 4+h
f = i
g =j

c-k = (100e+10f+g)-(100h+10i+j) = 400

Answer is A.
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The function f is defined as follows: for any 3-digit integer (written 16 May 2011, 20:41
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c-k = (x+4) * 100 + y*10 + z - ( x*100+ y*10 + z)

leaves 400. A.
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The function f is defined as follows: for any 3-digit integer (written 16 May 2011, 20:56
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400

as only powers of 2 can make it 16 times

so hundredth digit must be 4 more
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The function f is defined as follows: for any 3-digit integer (written 17 May 2011, 08:58
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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
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The function f is defined as follows: for any 3-digit integer (written 11 Mar 2015, 23:11
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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:
Show Spoiler
A

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The function f is defined as follows: for any 3-digit integer (written 19 Jul 2015, 05:57
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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!
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The function f is defined as follows: for any 3-digit integer (written 10 Nov 2018, 06:48
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fluke
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


©Grockit
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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)
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The function f is defined as follows: for any 3-digit integer (written 13 Dec 2018, 20:32
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pike
\(c = efg\)
\(k = lmn\)

\(f(c) = 2^e*3^f*5^g\)
\(f(k) = 2^l*3^m*5^n\)

\(f(c) = 16 * f(k)\)
\(2^e*3^f*5^g = 2^4 * 2^l*3^m*5^n\)

So, \(e = (4 + l), f = m, g = n\)

\(c - k = efg - lmn = (4+l)mn - lmn = 400\)

e.g k = 100, f(k) = 2, f(c) = 16*f(k) = 32 = 2^5, therefore c = 500, 500 - 100 = 400
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Can you pls explain the highlighted part?
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The function f is defined as follows: for any 3-digit integer (written 06 Oct 2024, 12:45
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