TITLE: If $2^{2013}-2^{2012}-2^{2011}+2^{2010} = k \cdot 2^{2010}$, find $k$
QUESTION [2 upvotes]: Hey this is just a question i was having fun with but couldnt solve for some reason. Would love if you can help me solve it thankyou!:
If $2^{2013}-2^{2012}-2^{2011}+2^{2010} = k\times2^{2010}$.  find k

REPLY [1 votes]: Well there is of course the option of simplifying via a calculator but if you want to do it by hand, first I would pull out the factor of $2^{2010}$ out of the left hand side to get
$$2^{2010}(2^3-2^2-2+1)=k\cdot 2^{2010}$$
Then simplify and divide both sides by $2^{2010}$ to get
$$\frac{2^{2010}\cdot 3}{2^{2010}}=k$$
Which yields
$$3=k$$

REPLY [0 votes]: Let $a=2^{2010}$ and just bring down some of the exponents:
$$2^{2013}-2^{2012}-2^{2011}+2^{2011} = 8a - 4a -2a +a$$
$$= 3a = 3\times 2^{2010}$$
So $k=3$.

REPLY [0 votes]: The knowledge you want to use:
$$x^{a+b}=x^ax^b$$
Hence you should be able to write everything in multiple of $2^{2010}$, by manipulating the coefficients to get $k$
Solution:
$$2^{2013} - 2^{2012} - 2^{2011} + 2^{2011}\\
=2^3 \cdot 2^{2010} - 2^2 \cdot 2^{2010}
=6 \cdot 2^{2010}$$

REPLY [0 votes]: We have
$$2^{x+2} - 2^{x+1} + 2^{x} = k\cdot 2^{x}$$
therefore
$$k = 2^{2} - 2^{1} + 1 = 3$$
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