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26 Feb 2019, 15:03
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
53% (03:01) correct
47%
(03:25)
wrong
based on 157
sessions
History
Date
Time
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Not Attempted Yet
GMATH practice exercise (Quant Class 16)
If x, y, and z are integers such that 67500 is divisible by (2^x)(3^y)(5^z) and (2^x)(3^y)(5^z) is NOT a multiple of 54, what is the maximum possible value of 3x+2y+z?
(A) 15
(B) 14
(C) 13
(D) 12
(E) less than 12
If x, y, and z are integers such that 67500 is divisible by (2^x)(3^y)(5^z) and (2^x)(3^y)(5^z) is NOT a multiple of 54, what is the maximum possible value of 3x+2y+z?
(A) 15
(B) 14
(C) 13
(D) 12
(E) less than 12
alldaytestprep
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Joined: 08 May 2018
Last visit: 24 Apr 2023
Posts: 98
Given Kudos: 1
Affiliations: All Day Test Prep
Location: United States (IL)
Schools: Booth '20 (A)
GMAT 1: 770 Q51 V49
GRE 1: Q167 V167
GPA: 3.58
Expert reply
26 Feb 2019, 15:59
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fskilnik
Step 1: Break up 67500 into it's prime factors and find out how many 2's, 3's, and 5's we have.
- 67500 = (2^2)(3^3)(5^4)
Step 2: Make sure our factors don't include a 54
- 54 = (3^3)(2)
- We want to use take out the MINIMUM value of prime factors possible(leave in the max possible factor) so that they no longer contain a 54
- This means we can take out two of our 2's or one of our 3's.
- Since 3 < 2^2 take out a 3.
Step 3: Max Factor = Whatever Prime Factors are left
- (2^2)(3^2)(5^4)
- therefore x =2, y =2, z =4
Step 4: Calculate 3x + 2y + z
- 6 + 4 + 4 = 14
Answer: B
General Discussion
27 Feb 2019, 08:08
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fskilnik
Very nice, arosman. Thank you for your contribution (and kudos)!
Our official solution follows:
\(? = \max \left( {3x + 2y + z} \right)\,\,\,\left( * \right)\)
\(x,y,z\,\,\mathop \ge \limits^{\left( * \right)} \,\,0\,\,\,{\rm{ints}}\,\,\,\left( {**} \right)\)
\(67500 = \underleftrightarrow {675 \cdot 100} = 25 \cdot 27 \cdot 4 \cdot 25 = {2^2} \cdot {3^3} \cdot {5^4}\)
\(54 = 2 \cdot 27 = 2 \cdot {3^3}\)
\(\left. \matrix{\\
{\mathop{\rm int}} = {{\,{2^2} \cdot {3^3} \cdot {5^4}\,} \over {{2^x} \cdot {3^y} \cdot {5^z}}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\left\{ \matrix{\\
\,0 \le x \le 2 \hfill \cr \\
\,0 \le y \le 3 \hfill \cr \\
\,0 \le z \le 4 \hfill \cr} \right. \hfill \cr \\
{\mathop{\rm int}} \ne {{\,{2^x} \cdot {3^y} \cdot {5^z}\,} \over {2 \cdot {3^3}}}\,\,\,\, \Rightarrow \,\,\,\,x = 0\,\,{\rm{or}}\,\,y < 3\,\,\left( {{\rm{or}}\,\,{\rm{both}}} \right)\,\, \hfill \cr} \right\}\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\rm{Take}}\,\,\,\left( {x,y,z} \right) = \left( {2,3 - 1,4} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 14\)
The correct answer is (B).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
