A to Z: Kids Problem

Algebra Level 2

X = a 1 ⋅ b 3 ⋅ c 5 ⋅ ⋅ ⋅ ⋅ z 51 Y = z 1 ⋅ y 3 ⋅ x 5 ⋅ ⋅ ⋅ ⋅ a 51 \begin{aligned}X&=&a^1\cdot b^3 \cdot c^5\cdot \cdot \cdot \cdot z^{51} \\ Y&=&z^1 \cdot y^3 \cdot x^5 \cdot \cdot \cdot \cdot a^{51} \end{aligned}

If a ⋅ b ⋅ c ⋅ d ⋅ ⋅ ⋅ ⋅ z = ( − 24389 ) 1 156 a \cdot b \cdot c \cdot d \cdot \cdot \cdot \cdot z=(-24389)^{\frac{1}{156}} , then find the value of X Y XY .


The answer is -29.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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2 solutions

Vaibhav Prasad
May 5, 2015

X = a 1 ⋅ b 3 ⋅ c 5 ⋅ ⋅ ⋅ ⋅ z 51 Y = z 1 ⋅ y 3 ⋅ x 5 ⋅ ⋅ ⋅ ⋅ a 51 ⇒ X Y = ( a 1 ⋅ b 3 ⋅ c 5 ⋅ ⋅ ⋅ ⋅ z 51 ) ( z 1 ⋅ y 3 ⋅ x 5 ⋅ ⋅ ⋅ ⋅ a 51 ) = a 52 ⋅ b 52 ⋅ c 52 ⋅ ⋅ ⋅ ⋅ z 52 ⇒ X Y = ( a ⋅ b ⋅ c ⋅ ⋅ ⋅ ⋅ z ) 52 X=a^{ 1 }\cdot b^{ 3 }\cdot c^{ 5 }\cdot \cdot \cdot \cdot z^{ 51 }\\ Y=z^{ 1 }\cdot y^{ 3 }\cdot x^{ 5 }\cdot \cdot \cdot \cdot a^{ 51 }\\ \\ \Rightarrow XY=(a^{ 1 }\cdot b^{ 3 }\cdot c^{ 5 }\cdot \cdot \cdot \cdot z^{ 51 })(z^{ 1 }\cdot y^{ 3 }\cdot x^{ 5 }\cdot \cdot \cdot \cdot a^{ 51 })=a^{ 52 }\cdot b^{ 52 }\cdot c^{ 52 }\cdot \cdot \cdot \cdot z^{ 52 }\\ \Rightarrow XY={ (a\cdot b\cdot c\cdot \cdot \cdot \cdot z) }^{ 52 }

Now, since

a ⋅ b ⋅ c ⋅ d ⋅ ⋅ ⋅ ⋅ z = ( − 24389 ) 1 156 ⇒ ( a ⋅ b ⋅ c ⋅ d ⋅ ⋅ ⋅ ⋅ z ) 52 = [ ( − 24389 ) 1 156 ] 52 ⇒ X Y = [ ( − 24389 ) 1 156 ] 52 = ( − 24389 ) 1 3 ⇒ X Y = − 29 a\cdot b\cdot c\cdot d\cdot \cdot \cdot \cdot z=(-24389)^{ \frac { 1 }{ 156 } }\\ \Rightarrow { (a\cdot b\cdot c\cdot d\cdot \cdot \cdot \cdot z) }^{ 52 }={ \left[ (-24389)^{ \frac { 1 }{ 156 } } \right] }^{ 52 }\\ \Rightarrow XY={ \left[ (-24389)^{ \frac { 1 }{ 156 } } \right] }^{ 52 }=(-24389)^{ \frac { 1 }{ 3 } }\\ \Rightarrow XY=\boxed { -29 }

24 Helpful 0 Interesting 2 Brilliant 0 Confused

Well, Vaibhav do you know any easy method for finding cube roots??????

Abhisek Mohanty - 6 years ago

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I don't, but it's not too hard to find the cube root of − 24389 -24389 .

( − 20 ) 3 = − 8000 (-20)^3 = -8000

( − 30 ) 3 = − 27000 (-30)^3 = -27000

The answer is between − 20 -20 and − 30 -30 , and much closer to − 30 -30 . For now, we can guess the answer is an integer. Since the units digit of − 24389 -24389 is 9 9 , the units digit of the cube root, when cubed, should also be 9 9 . The only such digit is 9 3 = 729 9^3 = 729 . Since − 29 -29 is a good candidate, we can multiply it out to confirm.

Zain Majumder - 2 years, 9 months ago

Nice man!!

Keegan Murphy - 3 years ago
Ahmed Obaiedallah
Jul 11, 2015

X = a 1 ⋅ b 3 ⋅ c 5 … z 51 Y = z 1 ⋅ y 3 ⋅ x 5 … a 51 \Large\begin{aligned}X&=&a^1\cdot b^3 \cdot c^5\ldots z^{51} \\ Y&=&z^1 \cdot y^3 \cdot x^5 \ldots a^{51} \end{aligned}

a ⋅ b ⋅ c ⋅ d … z = ( − 24389 ) 1 156 \Large a \cdot b \cdot c \cdot d \ldots z=(-24389)^{\frac{1}{156}} ,

X Y = ( a 1 ⋅ b 3 ⋅ c 5 … z 51 ) × ( z 1 ⋅ y 3 ⋅ x 5 … a 51 \Large XY=(a^{1}\cdot b^{3} \cdot c^{5}\ldots z^{51})\times(z^{1} \cdot y^{3} \cdot x^{5} \ldots a^{51}

= ( a ⋅ b ⋅ c ⋅ d … z ) 52 \Large=(a \cdot b \cdot c \cdot d \ldots z)^{52}

= ( ( − 24389 ) 1 156 ) 52 \Large=((-24389)^{\frac{1}{156}})^{52}

= ( ( − 24389 ) 1 3 \Large=((-24389)^{\frac{1}{3}}

= − 24389 3 = − 29 \LARGE=\sqrt[3]{-24389}=\boxed{\boxed{\color{#D61F06}{-29}}}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Good solution

Sai Ram - 5 years, 9 months ago

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