Powerful expressions
Put these expressions in order, from smallest to largest.
Put these expressions in order from smallest to largest:
a) $2^7+2$
b) $2\left(4^3\right)-2$
c) $8^2+4^3$
d) $8\times4\times2$
b) $2\left(4^3\right)-2$
c) $8^2+4^3$
d) $8\times4\times2$
This problem is adaped from the World Mathematics Championships
Answer: The correct order, from smallest to largest, is d, b, c, a.
Writing all the expressions in terms of 2
$\begin{align}2\left(4^3\right)-2 &= 2\left(\left(2^2\right)^3\right)-2\\
& = 2\left(2^6\right)-2\\
& = 2^1\times2^6-2\\
& = 2^7-2\\
\\
8^2+4^3&=\left(2^3\right)^2 + \left(2^2\right)^3\\
&=2^6+2^6\\
&=2\times2^6\\
&=2^7\\
\\
8\times4\times2 & = 2^3\times 2^2\times2\\
& = 2^{3+2+1}\\
&=2^6\end{align}$
$2^6$ $<$ $2^7-2$ $<$ $2^7$ $<$ $2^7+2$,
so $8\times4\times2<2\left(4^3\right)-2<8^2+4^3<2^7+2$
Using the 8 times table to work out the values of the expressions
$2^2 = 4$ and $2^3 = 8$, and $2^7=2^{3+3+1}=2^3\times2^3\times2^1$.
So $2^7 = 8\times8\times2 = 64\times2=128,$
so $2^7+2=128+2=130.$
$2\left(4^3\right) = 2\times 4\times 4 \times 4 = 8\times 16 = 8\times2 \times 2$ again.
So $2\left(4^3\right) = 128,$
so $2\left(4^3\right)-2 = 128-2=126.$
$8^2 + 4^3 = 64 + 4\times4\times4$. We can use $8$s again to work out $4\times4\times4$:
$4\times4\times4= 4\times (2\times2)\times 4 = \left(4\times2\right)\times\left(2\times4\right) = 8\times8=64.$
So $8^2+4^3 = 64+64=128.$
And $8\times4\times2 = 8\times8=64.$
Clearly $64<126<128<130$, so $8\times4\times2<2\left(4^3\right)-2<8^2+4^3<2^7+2$