Take three consecutive positive integers and cube them. Add up the digits in each of the three results, and add again until you’ve reached a single digit for each of the three numbers.
For example:
46^3= 97336; 9 + 7 + 3 + 3 + 6 = 28; 2 + 8 = 10; 1 + 0 = 1
47^3= 103823; 1+ 0 + 3 + 8 + 2 + 3 = 17; 1 + 7 = 8
48^3= 110592; 1 + 1 + 0 + 5 + 9 + 2 = 18; 1 + 8 = 9
Arrange the 3 digits in ascending order.
ALWAYS 189 !
Please explain why.
Source: Kendall and Thomas’ Mathematical Puzzles for the Connoisseur (1971)
If you have 3 consecutive numbers, one will be a multiple of 3. One of the others will be one less, and the other one more than a multiple of 3. Call them 3a-1, 3b, 3c+1.
(3a-1)^3 = 27a^3 - 27a^2 + 9a - 1 which is one less than a multiple of 9m so the process ends with 8
(3b)^3 is a multiple of 9, so the process ends with 9
(3c+1)^3 = 27c^3 + 27c^2 + 9c + 1 which is one more than a multiple of 9m so the process ends with 1
If you put them in ascending order you get 189.
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Posted by Jer
on 2023-05-05 11:31:58 |
Solution
