Difference between revisions of "2001 OIM Problems/Problem 1"
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* All digits of <math>n</math> are greater than 1. | * All digits of <math>n</math> are greater than 1. | ||
| − | + | * Whenever four digits of <math>n</math> are multiplied, a divisor of <math>n</math> is obtained. | |
| − | Show that for each natural number < | + | 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 | ||
Latest revision as of 03:08, 14 December 2023
Problem
We say that a natural number 
is "çharrúa" if it simultaneously satisfies the following conditions:
- All digits of
are greater than 1.
- Whenever four digits of are multiplied, a divisor of is obtained.
Show that for each natural number 
there is a çharrúa number with more than digits.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
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