Question

Find the easiest way to solve
3^2008 (10^2013 + 5^2012 + 2^2011)
------------------------------------
5^2012 (6^2010 + 3^2009 + 2^2008)

Answer 1

@zzr0ck3r

Answer 2

Don't need to find the final answer, just how is the easiest way to solve it.

Answer 3

@ganeshie8

Answer 4

@princeharryyy

Answer 5

I don't know what are you solving for. If you are considering adding them. Sorry. I don't think you can. But, remainder is possible.

Answer 6

I don't know what are you solving for. If you are considering adding them. Sorry. I don't think you can. But, remainder is possible.

Answer 7

This was the question that my teacher gave to me. You don't need to solve it anyway. You just should find what is 'the easiest way' to solve it.

Answer 8

Note: 10=2*5 and 6=2*3

Answer 9

How about '+' symbols?

Answer 10

You're going to have to use the distributive property too.

Answer 11

$$ 3^{2008} (10^{2013} + 5^{2012} + 2^{2011})= 3^{2008} ((2•5)^{2013} + 5^{2012} + 2^{2011})$$

Answer 12

can we divide numerator with denominator?

Answer 13

hmmm... this is a tough one

Answer 14

$$\huge\dfrac{ 3^{2008} ((2•5)^{2013} + 5^{2012} + 2^{2011})}{5^{2012} ((2•3)^{2010} + 3^{2009} + 2^{2008})}$$

$$\huge = \dfrac{ 3^{2008} (2^{2013}•5^{2013} + 5^{2012} + 2^{2011})}{5^{2012} (2^{2010}•3^{2010} + 3^{2009} + 2^{2008})}$$

Answer 15

I still get no clue..
why if we distribute it like that, we can solve it faster?

Answer 16

I'm looking for common factors...
Perhaps the prime factorization of 2008, 2009, 2010, 2011, 2012, and 2013 will show some pattern.

Answer 17

I got it. Tq very much!

Answer 18

Please share the answer...

Answer 19

all I could find was it's approximately equal to 40/9

Answer 20

Was it really division? @Kevin

Answer 21

I was on mobile and I thought it was a series of something ------- in between . LOL

Answer 22

easiest way>> use logarithms to solve.

Answer 23

Sorry, I write the question wrong. The true question like this :
3^2008 (10^2013 + 5^2012 x 2^2011)
------------------------------------
5^2012 (6^2010 + 3^2009 x 2^2008)

So the answer would be like this :
3^2008 ((2 x 5)^2013 + 5^2012 x 2^2011)
---------------------------------------- =
5^2012 (2 x 3)^2010 + 3^2009 x 2^2008)

3^2008 x 2^2013 x 5^2013 + 3^2008 x 5^2012 x 2^2011
------------------------------------------------------ =
5^2012 x 2^2010 x 3^2010 + 5^2012 x 3^2009 x 2^2008

3^2008 x 2^2008 x 5^2012 (2^5 x 5 + 2^3)
----------------------------------------- =
3^2008 x 2^2008 x 5^2012 (3^2 x 2^2 + 3)

Until this, we know what is the next step :D
Pls correct my answer if you find something wrong.

Answer 24

Yes, it's division @princeharryyy
It's algebra contains division in it

Answer 25

You are doing it okay. Cancel out the common terms.

Answer 26

(2^5 x 5 + 2^3)
-------------- = (32*5 +8)/(36+3) = 168/39
(3^2 x 2^2 + 3)

Answer 27

168/39 = 56/13

Answer 28

yeah, that's my final answer :D
Sorry If I write the question wrong and not clearly

Answer 29

Tq for your help

Answer 30

It's okay. All I thought was that there were many terms. And the only possible question would have been to find the remainder like your earlier problem. For these kind of division problems you always take out the common terms. and if you can't find anything common you can find individual values using logarithms and try to find the answer.

Answer 31

$$\large\dfrac{3^{2008} (10^{2013} + 5^{2012} 2^{2011})}{5^{2012} (6^{2010} + 3^{2009} 2^{2008})}$$

$$\large=\dfrac{3^{2008} ((2\cdot5)^{2013} + 5^{2012} 2^{2011})}{5^{2012} ((2\cdot3)^{2010} + 3^{2009} 2^{2008})}$$

$$\large=\dfrac{3^{2008} (2^{2013}\cdot5^{2013} + 5^{2012} 2^{2011})}{5^{2012} (2^{2010}\cdot3^{2010} + 3^{2009} 2^{2008})}$$

$$\large=\dfrac{3^{2008} 2^{2011} (5^{2013} 2^2 + 5^{2012} )}{5^{2012} 2^{2008} (2^{2} \cdot3^{2010} + 3^{2009} )}$$

$$\large=\dfrac{3^{2008} 2^{2011} 5^{2012} (5 \cdot 2^2 +1 )}{5^{2012} 2^{2008} 3^{2009} (2^{2} \cdot3 + 1 )}$$

$$\large=\dfrac{3^{2008} 2^{2011} \cancel{5^{2012}} (5 \cdot 2^2 +1 )}{\cancel{5^{2012}} 2^{2008} 3^{2009} (2^{2} \cdot3 + 1 )}$$

$$\large=\dfrac{3^{2008} 2^{2011} (5 \cdot 2^2 +1 )}{ 2^{2008} 3^{2009} (2^{2} \cdot3 + 1 )}$$

$$\large=\dfrac{ 2^{2011} (5 \cdot 2^2 +1 )}{ 2^{2008} 3 ( 2^{2} \cdot3 + 1 )}$$

$$\large=\dfrac{ 2^{3} (5 \cdot 2^2 +1 )}{ 3 ( 2^{2} \cdot3 + 1 )}$$

$$\large=\dfrac{ 8 (5 \cdot 4 +1 )}{ 3 ( 4 \cdot3 + 1 )}$$

$$\large=\dfrac{ 8 (21 )}{ 3 ( 13 )}=56/13$$

Answer 32

Yeah, that's it
It's the correct question...

Answer 33

Good job :-)