An oldie

[Rob625]

An old puzzle to start off a new board.

If you cube each digit of the number 153 and add, you get back to 153: 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153.

Which other 3-digit numbers have this property?

[ReformedCharacter]

370
371
407

RC

[Gengulphus]

An old puzzle to start off a new board.

If you cube each digit of the number 153 and add, you get back to 153: 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153.

Which other 3-digit numbers have this property?

Not certain what good a spoiler separator will do in this see-20-posts-at-once environment, but let's have one anyway...

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We want solutions to:

100A + 10B + C = A^3 + B^3 + C^3

with A, B and C in the range 0-9, and we might or might not be interested in solutions with leading zeros, i.e. with A=0.

Rearranging, that is:

(100A-A^3) + (10B-B^3) + (C-C^3) = 0

So we want to take one number from each of the 2nd to 4th columns in the following table, such that the three numbers we've taken add up to zero:

Digit    100A-A^3    10B-B^3     C-C^3
--------------------------------------
  0          0           0          0
  1         99           9          0
  2        192          12         -6
  3        273           3        -24
  4        336         -24        -60
  5        375         -75       -120
  6        384        -156       -210
  7        357        -273       -336
  8        288        -432       -504
  9        171        -639       -720

If B is one of 0-3, we're looking for a positive number in the 2nd column that is 0, 3, 9 or 12 less in magnitude than a negative number in the 4th column. It is easy to work through the numbers in the 2nd column, looking to see whether the 4th column contains suitable numbers: the results are that 0 and 336 work and the others don't. This leads to the solutions:

000 = 0^3 + 0^3 + 0^3 (if one is interested in solutions with leading zeros)
001 = 0^3 + 0^3 + 1^3 (if one is interested in solutions with leading zeros)
407 = 4^3 + 0^3 + 7^3

Otherwise B is one of 4-9, and the numbers taken from the 3rd and 4th columns are both negative. So their negations B^3-10B and C^3-C are both positive and add up to the number taken from the 2nd column, which is at most 384. We can therefore rule out 8 and 9 as possibilities for B and C, as they lead to one or both of those two positive numbers being > 384. Writing out the addition table for the remaining possible values of B^3-10B and C^3-C and asterisking the ones that are possible values of A, we get:

+   |    0     6    24    60   120   210   336
----+------------------------------------------
 24 |   24    30    48    84   144   234   360
 75 |   75    81    99*  135   195   285   411
156 |  156   162   180   216   276   366   492
273 |  273*  279   297   333   393   483   609

The two asterisked entries lead to the remaining three solutions:

153 = 1^3 + 5^3 + 3^3 (which is the solution given in the problem)
370 = 3^3 + 7^3 + 0^3
371 = 3^3 + 7^3 + 1^3

So the other 3-digit numbers that have the property are 370, 371 and 407, and 000 and 001 if you allow leading zeros.

Gengulphus

[ReformedCharacter]

Bravo Gengulphus.

RC

[NomoneyNohoney]

Bravo Gengulphus.

RC

And if that doesn't reinforce the need for recs, nothing will!
Bravo Gengulphus 2