Difference between revisions of "1999 OIM Problems/Problem 1"
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<math>1^3 = 1^2 , 8^3 \neq 8^2, 9^3 = 27^2, 10^3 \neq 64^2, 8^3 \neq 125^2, 13^3 \neq 256^2 , 10^3 \neq 343^2 , 8^3 \neq 512^2 , </math> and <math>18^2 \neq 729^2</math>. | <math>1^3 = 1^2 , 8^3 \neq 8^2, 9^3 = 27^2, 10^3 \neq 64^2, 8^3 \neq 125^2, 13^3 \neq 256^2 , 10^3 \neq 343^2 , 8^3 \neq 512^2 , </math> and <math>18^2 \neq 729^2</math>. | ||
So the only integers that satisfy this condition are <math>1</math> and <math>27</math> | So the only integers that satisfy this condition are <math>1</math> and <math>27</math> | ||
| − | + | ~Archieguan | |
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe14.htm | https://www.oma.org.ar/enunciados/ibe14.htm | ||
Revision as of 00:50, 23 October 2024
Problem
Find all positive integers that are less than 1000 and satisfy the following condition: the cube of the sum of their digits is equal to the square of that integer.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Insight: Every number that satisfies this must be a cube itself
Proof/reasoning: let the sum of digits be 
and the original number be 
. Then 
. If weren’t a cube, neither would 
, but it is. Therefore, is a cube.
Now we list out all cubes that are smaller than 
and 
.
So the only integers that satisfy this condition are 
and 
~Archieguan
