If 3^(3 + 6 + 9 + … + 3x) = (37/999)^(-66), what is the value of x?

Question

If  3^{(3 + 6 + 9 + … + 3x)} = (\frac{37}{999})^{(-66)} , what is the value of x?

A. 3
B. 6
C. 7
D. 9
E. 11

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chetan2u · Accepted Answer

999 has to be a multiple of 37 to simplify this question, and if you realize this, this question is a cake walk.

3^{(3 + 6 + 9 + … + 3x)} = (\frac{37}{999})^{(-66)}=\frac{37}{37*27}^{-66}=(3^3)^{66}=3^{198}

Equating the powers..

3+6+9+...3x=198.......3(1+2+..x)=198 ...
 1+2+...+x=66.........\frac{x(x+1)}{2}=66......x(x+1)=66*2=11*12....x=11

E

ayushkumar22941 · Answer

first of all 999 can be written as 999=3*3*3*37
this will simplify right side of equation to 3power198 
then we equate left and right side power as base 3 is common 
lest side forms an AP and using formula for sum of ap we get value of n =11 so e is correct choice

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If 3^(3 + 6 + 9 + … + 3x) = (37/999)^(-66), what is the value of x?

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If 3^(3 + 6 + 9 + … + 3x) = (37/999)^(-66), what is the value of x?

Prep Toolkit

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Perfect 805 Score Debrief by Julia

My Rewards