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

2 Answers
Aug 11, 2016

Answer:

#729#

Explanation:

Let #n# be this number.

Then #n = m^6 = (m^2)^3 = (m^3)^2# so a candidate is #m = 3# because

#n = 729 = (3^2)^3 = (3^3)^2#

Sep 2, 2016

Answer:

#(3^2)^3 =9^3 = 729#

#(3^3)^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 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 ...#

#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#