Find all three-digit numbers that are equal to the sum of the cubes of the digits.

Guest May 31, 2020

#2**0 **

As pointed out, a part of the problem you have raised refers to Armstrong numbers / narcissistic numbers, i.e., nn -digit numbers equal to the sum of nn -th power of its digits. Besides that, there is nothing really special about such numbers. For n=3n=3 , there are only 44 such numbers, 153,370,371153,370,371 and 407407 . There is no analytical means to solve those equations exactly, thus one has to rely on trial-and-error or brute force.

The other part, where a 4-digit number is equal to sum of third powers of its digits, does not have any solution. Consider, as you showed:

a3+b3+c3=100a+10b+ca3+b3+c3=100a+10b+c

Adding, 10001000 on both sides, we have:

103+a3+b3+c3=1000+100a+10b+c103+a3+b3+c3=1000+100a+10b+c

Clearly, for any d∈[1,9]d∈[1,9] , we have:

103+a3+b3+c3≤1000d+100a+10b+c103+a3+b3+c3≤1000d+100a+10b+c

And thus:

d3+a3+b3+c3<1000d+100a+10b+cd3+a3+b3+c3<1000d+100a+10b+c

Conclusion: no solutions exist.

Hope this helps, whymenotsmart^m^, although guest's answer is alternable:)

whymenotsmart May 31, 2020