1999 OIM Problems/Problem 1: Difference between revisions
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== Solution == | == Solution == | ||
Insight: Every number that satisfies this must be a cube itself | |||
Proof/reasoning: let the sum of digits be <math>r</math> and the original number be <math>n</math>. Then <math>r^3 = n^2</math>. If <math>n</math> weren’t a cube, neither would <math>n^2</math>, but it is. Therefore, <math>n</math> is a cube. | |||
Now we list out all cubes that are smaller than <math>1000</math> | |||
<math>1,8,27,64,125,256,343,512,729</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> | |||
~Archieguan | |||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe14.htm | https://www.oma.org.ar/enunciados/ibe14.htm | ||
Latest 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
