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GMATH Teacher P
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Joined: 12 Oct 2010
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If x, y, and z are integers such that 67500 is divisible by (2^x)(3^  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 52% (03:20) correct 48% (03:11) wrong based on 33 sessions

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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

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
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Re: If x, y, and z are integers such that 67500 is divisible by (2^x)(3^  [#permalink]

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fskilnik wrote:
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

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

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GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If x, y, and z are integers such that 67500 is divisible by (2^x)(3^  [#permalink]

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fskilnik wrote:
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

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.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net Re: If x, y, and z are integers such that 67500 is divisible by (2^x)(3^   [#permalink] 27 Feb 2019, 08:08
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