Why does $(2^{20}+2^{20}+2^{20}+2^{21})=5\cdot 2^{20}$?

Question

I did this question on artofproblemsolving.com and I do not understand the solution.

Why do I have $5 \cdot 2^{20}$? Can anyone explain?

Answer 1

The basic rules you need to use are these: $$ \alpha a^n+\beta a^n+\cdots+\omega a^n=(\alpha+\beta+\cdots+\omega)\cdot a^n\tag{1} $$ and $$ a^n\cdot a^m=a^{n+m}.\tag{2} $$ Now let's consider what you have in the numerator: $$ 2^{20}+2^{20}+2^{20}+2^{21}=2^{20}+2^{20}+2^{20}+\underbrace{2\cdot 2^{20}}_{\text{by $(2)$}}=\underbrace{(1+1+1+2)\cdot 2^{20}}_{\text{by $(1)$}}=\color{blue}{5}\cdot 2^{20}. $$ Does that make sense? Most of these kinds of "tricky" problems come down to seeing how to manipulate exponents effectively. For example, consider the following "tricky" problem:

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Problem: How many digits are there in the product $8^{15}\cdot 5^{37}$ [deduce by manual operations]?

Solution. Notice the following (try to figure it out first):

$$8^{15}\cdot 5^{37}=(2^3)^{15}\cdot 5^{37}=2^{45}\cdot 5^{37}=2^8\cdot2^{37}\cdot 5^{37}=2^8\cdot 10^{37}=256\times 10^{37}.$$ Thus, the number has $3+37=40$ digits.

As you can see, many of these problems simply rely on being able to manipulate quantities/terms effectively. The more problems you do, the better you will get (as always).

Answer 2

$2^{21}=2\cdot 2^{20}$ hence $2^{20}+2^{20}+2^{20}+2^{21}=5\cdot 2^{20}$ and $5\cdot 2^{20}/2^{17}=5\cdot 2^{20-17}=5\cdot 8=40$.

Answer 3

$$(2^{20}+2^{20}+2^{20}+2^{21})/ 2^{17}=$$ $$2^{20}(1+1+1+2)/2^{17} = 5\cdot2^{20}/2^{17}$$

Answer 4

we have $(2^{20}+2^{20}+2^{20}+2^{20}\cdot 2)/2^{17}=\frac{2^{20}}{2^{17}}(1+1+1+2)=5\cdot 2^3=40$

Answer 5

Hint 1: $\ 2^{21}=2^{20+1}=2\cdot 2^{20}$.

Hint 2: $\ 2^{20}+2^{20}=2 \cdot 2^{20}$

Spoiler:

$2^{20}+2^{20}+2^{20}+2^{21}=3 \cdot 2^{20} + 2 \cdot 2^{20}= 5 \cdot 2^{20}\implies \displaystyle\frac{5\cdot 2^{20}}{2^{17}}= 5\cdot\displaystyle\frac{2^{20}}{2^{17}}=5\cdot 2^{20-17}=5\cdot 2^3 = 40$

Answer 6

\begin{align} 2^{20}+2^{20}+2^{20}+2^{21}&=\phantom{1\cdot{}}2^{20}+ \phantom{1\cdot{}}2^{20}+ \phantom{1\cdot{}}2^{20}+2\cdot2^{20}\\ &=1\cdot2^{20}+1\cdot2^{20}+1\cdot2^{20}+2\cdot2^{20}\tag{1}\\ &=(1+1+1+2)\cdot2^{20}\tag{2}\\ &=5\cdot2^{20} \end{align} where in equation $(1)$ I used the fact that 1 times anything is itself, and in equation $(2)$ I used the distributive property.