Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 OIM Problems/Problem 1"

(Created page with "== Problem == We say that a natural number <math>n</math> is "''çharrúa''" if it simultaneously satisfies the following conditions: * All digits of <math>n</math> are grea...")
 
(solution)
 
(One intermediate revision by one other user not shown)
Line 5: Line 5:
 
* All digits of <math>n</math> are greater than 1.
 
* All digits of <math>n</math> are greater than 1.
  
−
<math> Whenever four digits of </math>n<math> are multiplied, a divisor of </math>n<math> is obtained.
+
* Whenever four digits of <math>n</math> are multiplied, a divisor of <math>n</math> is obtained.
  
−
Show that for each natural number </math>k<math> there is a ''çharrúa'' number with more than </math>k$ digits.
+
Show that for each natural number <math>k</math> there is a ''çharrúa'' number with more than <math>k</math> digits.
  
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
Clearly if we show that there are infinite such numbers, then the criterion is satisfied. Thus we create such a number. Define
 +
<cmath>X=22223232</cmath>
 +
Notice that <math>X</math> clearly has <math>2\cdot2\cdot2\cdot2=16</math> as a factor and also has <math>3\cdot3=9</math> as a factor by the sum of the digits; therefore, this number is <i>çharrúa</i>.
 +
 
 +
Next, let <math>a</math> be some natural number divisible by <math>9</math>, and add <math>a</math> <math>2</math>'s to the beginning of <math>X</math> as digits. No additional possibilities for the product of four digits are introduced, and the resulting number is divisible by <math>9</math> by construction as well as <math>16</math> from before, and since <math>a</math> can be any positive multiple of <math>9</math>, the conclusion follows.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==

Latest revision as of 19:30, 10 April 2025

Problem

We say that a natural number $n$ is "çharrúa" if it simultaneously satisfies the following conditions:

  • All digits of $n$ are greater than 1.
  • Whenever four digits of $n$ are multiplied, a divisor of $n$ is obtained.

Show that for each natural number $k$ there is a çharrúa number with more than $k$ digits.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Clearly if we show that there are infinite such numbers, then the criterion is satisfied. Thus we create such a number. Define \[X=22223232\] Notice that $X$ clearly has $2\cdot2\cdot2\cdot2=16$ as a factor and also has $3\cdot3=9$ as a factor by the sum of the digits; therefore, this number is çharrúa.

Next, let $a$ be some natural number divisible by $9$, and add $a$ $2$'s to the beginning of $X$ as digits. No additional possibilities for the product of four digits are introduced, and the resulting number is divisible by $9$ by construction as well as $16$ from before, and since $a$ can be any positive multiple of $9$, the conclusion follows.

~ eevee9406

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_OIM_Problems/Problem_1&oldid=246515"