fskilnik@GMATH 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

$$? = \max \left( {3x + 2y + z} \right)\,\,\,\left( * \right)$$
$$x,y,z\,\,\mathop \ge \limits^{\left( * \right)} \,\,0\,\,\,{\rm{ints}}\,\,\,\left( {**} \right)$$

$$67500 = \underleftrightarrow {675 \cdot 100} = 25 \cdot 27 \cdot 4 \cdot 25 = {2^2} \cdot {3^3} \cdot {5^4}$$
$$54 = 2 \cdot 27 = 2 \cdot {3^3}$$

$$\left. \matrix{
{\mathop{\rm int}} = {{\,{2^2} \cdot {3^3} \cdot {5^4}\,} \over {{2^x} \cdot {3^y} \cdot {5^z}}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\left\{ \matrix{
\,0 \le x \le 2 \hfill \cr
\,0 \le y \le 3 \hfill \cr
\,0 \le z \le 4 \hfill \cr} \right. \hfill \cr
{\mathop{\rm int}} \ne {{\,{2^x} \cdot {3^y} \cdot {5^z}\,} \over {2 \cdot {3^3}}}\,\,\,\, \Rightarrow \,\,\,\,x = 0\,\,{\rm{or}}\,\,y < 3\,\,\left( {{\rm{or}}\,\,{\rm{both}}} \right)\,\, \hfill \cr} \right\}\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\rm{Take}}\,\,\,\left( {x,y,z} \right) = \left( {2,3 - 1,4} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 14$$


The correct answer is (B).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

67500 = (2^2)*(3^3)*(5^4), which corresponds with (2^x)(3^y)(5^z). Thus 2 is divisible by x (i.e. x = 1 or x = 2), 3 is divisible by y (i.e. y = 1, y = 2, or y = 3), and 4 is divisible by z (i.e. z = 1, z = 2, z = 3, or z = 4).

54 = (2^1)*(3^3). Since (2^x)(3^y)(5^z) is not a multiple of 54, then x is not a multiple of 1 and/or y is not a multiple of 3. Since x must be a multiple of 1 (all integers are multiples of 1), then y is not a multiple of 3. The biggest value of x is when x = 2; the largest possible value of y that is not a multiple of 3 is when y = 2; and the largest possible value of z is when z = 4.

Thus the largest possible value of 3x+2y+z = (3*2) + (2*2) + (4) = 6 + 4 + 4 = 14

Therefore the answer is (D).