Which 33 digit number is both a square and a cube?

2 Answers
Aug 11, 2016

729729

Explanation:

Let nn be this number.

Then n = m^6 = (m^2)^3 = (m^3)^2n=m6=(m2)3=(m3)2 so a candidate is m = 3m=3 because

n = 729 = (3^2)^3 = (3^3)^2n=729=(32)3=(33)2

Sep 2, 2016

(3^2)^3 =9^3 = 729(32)3=93=729

(3^3)^2 = 27^2 = 729(33)2=272=729

Explanation:

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

1^3 = 1^2 = 1 = 1^613=12=1=16 but it is a one digit number
2^3 = 823=8
3^3 = 2733=27
4^3 = 64 = 8^2 = 2^643=64=82=26- a two digit number
5^3 = 12553=125
6^3 = 21663=216
7^3 = 34373=343
8^3 = 51283=512
9^3 = 729 = 27^2 = 3^693=729=272=36
10^3 = 1000103=1000

There are 3 cubes which are also square numbers, but only 729729 is a three digit number. The next would be 4^6 =4096= 16^3 = 64^246=4096=163=642 a four-digit number.

Note that 1,4 and 91,4and9 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 ...

1^2 = 1
8^2=64
27^2 =729" "larr a three-digit number
64^2 =4096

However, if we look at the indices:

(x^2)^3 = (x^3)^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