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16 May 2011, 19:14
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (02:13) correct
42%
(02:20)
wrong
based on 431
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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
(A) 400
(B) 320
(C) k/(16c)
(D) 40
(E) Cannot be determined
©Grockit
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
\(k = lmn\)
\(f(c) = 2^e*3^f*5^g\)
\(f(k) = 2^l*3^m*5^n\)
\(f(c) = 16 * f(k)\)
\(2^e*3^f*5^g = 2^4 * 2^l*3^m*5^n\)
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
General Discussion
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.
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.
