Four digit cubes
Number Theory
Level
4
Find all four digits numbers which are equal to the cube of the sum of its digits.
Write you answer as the sum of these numbers.
The answer is 10745.
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1 solution
We are looking for 4 digit numbers, so those must be between 999 < X < 10000.
Let s(X) be the sum of X's digits, then
999^(1/3) < s(X) < 10000^(1/3)
9 < s(X) < 22
We can directly check all 12 number from 10 to 21, but we can cut on some brute forcing for multiples of 3.
If X is a multiple of 3, then X must also be a multiple of 9 (the factors of 3 is tripled making X a multiple of 27, but 27 is one of the multiples of 9). Thus, s(X) must be divisible by 9, too (check out wiki pages on divisibility rules) and s(X) = 18 is the only possibility within allowed range. Checking, we have 18³ = (2 × 9)³ = 2³ × 9³ = 8 × 729 = 729 × (9 – 1) = 81² – 729 = 6561 – 729 = 5832 with s(18) = 5 + 8 + 3 + 2 = 18 ✅
Now we're only left with 8 other numbers to explore if the number X is indeed equal to (s(X))³.
10³ = 1000 ≠ (1 + 0 + 0 + 0)³ = 1³
11³ = 1331 ≠ (1 + 3 + 3 + 1)³ = 8³
13³ = 2197 ≠ (2 + 1 + 9 + 7)³ = 19³
14³ = 2744 ≠ (2 + 7 + 4 + 4)³ = 17³
16³ = 4096 ≠ (4 + 0 + 9 + 6)³ = 19³
17³ = 4913 = (4 + 9 + 1 + 3)³ = 17³ ✅
19³ = 6859 ≠ (6 + 8 + 5 + 9)³ = 28³
20³ = 8000 ≠ (8 + 0 + 0 + 0)³ = 8³
Answer
= 18³ + 17³
= 5832 + 4913
= 10745