# Which 3 digit number is both a square and a cube?

Aug 11, 2016

$729$

#### Explanation:

Let $n$ be this number.

Then $n = {m}^{6} = {\left({m}^{2}\right)}^{3} = {\left({m}^{3}\right)}^{2}$ so a candidate is $m = 3$ because

$n = 729 = {\left({3}^{2}\right)}^{3} = {\left({3}^{3}\right)}^{2}$

Sep 2, 2016

${\left({3}^{2}\right)}^{3} = {9}^{3} = 729$

${\left({3}^{3}\right)}^{2} = {27}^{2} = 729$

#### Explanation:

One way to approach this is to consider the cubes and see which meet the other conditions.

${1}^{3} = {1}^{2} = 1 = {1}^{6}$ but it is a one digit number
${2}^{3} = 8$
${3}^{3} = 27$
${4}^{3} = 64 = {8}^{2} = {2}^{6}$- a two digit number
${5}^{3} = 125$
${6}^{3} = 216$
${7}^{3} = 343$
${8}^{3} = 512$
${9}^{3} = 729 = {27}^{2} = {3}^{6}$
${10}^{3} = 1000$

There are 3 cubes which are also square numbers, but only $729$ is a three digit number. The next would be ${4}^{6} = 4096 = {16}^{3} = {64}^{2}$ a four-digit number.

Note that $1 , 4 \mathmr{and} 9$ are the first three square numbers,
Their cubes are therefore both squares and cubes.
We could also have started with integers which are cubes, and then square them:

The cubes are:$1 , 8 , 27 , 64 \ldots$

${1}^{2} = 1$
${8}^{2} = 64$
${27}^{2} = 729 \text{ } \leftarrow$ a three-digit number
${64}^{2} = 4096$

However, if we look at the indices:

${\left({x}^{2}\right)}^{3} = {\left({x}^{3}\right)}^{2} = {x}^{6}$

Find the numbers which are sixth powers and have $3$ digits:

${1}^{6} = 1$
${2}^{6} = 64$
${3}^{6} = 729$
${4}^{6} = 4096$