The impossible equation

If { x = 20082008200820082008...20082008 2008 repeated 2018 times y = 20072007200720072007...20072007 2007 repeated 2018 times \begin{cases} x= \underbrace{20082008200820082008...20082008}_{\text{2008 repeated 2018 times}} \\ y=\underbrace{20072007200720072007...20072007}_{\text{2007 repeated 2018 times}} \end{cases} then find the value of 2007 x 2008 y 2007x-2008y .


The answer is 0.

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3 solutions

Chew-Seong Cheong
Jan 16, 2018

x = 200820082008...2008 2008 repeated 2018 times = 2018 ( 100010001...0001 0001 repeated 2018 times ) = 2018 a x = \underbrace{200820082008...2008}_{\text{2008 repeated 2018 times}} = 2018{\color{#3D99F6}(\underbrace{100010001...0001}_{\text{0001 repeated 2018 times}})} = 2018\color{#3D99F6}a , where a = 100010001...0001 0001 repeated 2018 times ) \color{#3D99F6} a = \underbrace{100010001...0001}_{\text{0001 repeated 2018 times}}) . Similarly, y = 2007 a y=2007a . Then 2007 x 2008 y = 2007 ( 2008 a ) 2008 ( 2007 a ) = 0 2007x-2008y = 2007(2008a) - 2008(2007a) = \boxed{0} .

Anuj Shikarkhane
Jan 21, 2018

x = 2008 ( 1 0 2018 + 1 0 2017 + 1 0 2016 + + 1 0 0 ) x = 2008(10^{2018}+10^{2017}+10^{2016}+\cdots +10^0)

y = 2007 ( 1 0 2018 + 1 0 2017 + 1 0 2016 + + 1 0 0 ) y = 2007(10^{2018}+10^{2017}+10^{2016}+\cdots +10^0)

Let's assign the letter P to 1 0 2018 + 1 0 2017 + 1 0 2016 + + 1 0 0 10^{2018}+10^{2017}+10^{2016}+\cdots +10^0

Therefore, x = 2008 P x = 2008P and y = 2007 P y=2007P

2007 x 2008 y = 2007 × 2008 P 2008 × 2007 P 2007x - 2008y= 2007 \times 2008P - 2008 \times 2007P

= 0 \boxed{0} .

Fantastic metbod

Ankita Das - 3 years, 4 months ago
Subhadeep Paul
Jan 17, 2018

x=2008(1+10^4+10^8+.... to 2018 terms) Therefore x=2008((10^4)^2018 -1)÷(10^4-1) Similarly y=2007*((10^4)^2018-1)÷(10^4-1) So we get, 2007x=2008y Therefore ans=0

Mast kiya bhai

Koulick Sadhu - 3 years, 4 months ago

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