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Perhaps there is a solution in a number system other than base 10. (I'm too lazy to pursue such a proof out of simple curiosity.) From the problem statement,
by DrScump 11y ago
Perhaps there is a solution in a number system other than base 10. (I'm too lazy to pursue such a proof out of simple curiosity.)
From the problem statement, you know that 0 and 9 exist, so that eliminates binary through Base 9. But perhaps this hints at a more broad question: can 9000 be arrived at in any numeric base (> 10)?
- tzs 11y agoThe sum of digits argument generalizes to all bases. Let the base be b. Then the sum of the unit column digits must be b. The sum of the b's column digits must be b-1, as must be the sum of the b^2's column digits. Finally, the sum of the b^3's column digits must be 8. That gives us a total sum of all the digits of 3b+6. Summing rows instead of columns gives us 1+2+3+4 in each row, and there are 4 rows, so that's a total of 40. Thus, we must have 3b+6 = 40. That has no solution in integers.