The residents of Octingham have 8 fingers. Instead of counting in base ten, they count in base eight: the digits of their numbers represent ones, eights, sixty-fours, two-hundred-and-fifty-sixes, etc instead of ones, tens, hundreds, thousands, etc.
For example, a residents of Octingham would say 12, 22 and 52 instead of our usual numbers 10, 18 and 42.
Today's number is what a resident of Octingham would call 11 squared (where the 11 is also written using the Octingham number system).
In bases 3 to 9, the number 112 is: \(11011_3\), \(1300_4\), \(422_5\), \(304_6\), \(220_7\), \(160_8\), and \(134_9\). In bases 3, 4, 6, 8 and 9, these representations contain no digit 2.
There are two 3-digit numbers that contain no 2 in their representations in all the bases between 3 and 9 (inclusive). Today's number is the smaller of these two numbers.
In base 2, 1/24 is 0.0000101010101010101010101010...
In base 3, 1/24 is 0.0010101010101010101010101010...
In base 4, 1/24 is 0.0022222222222222222222222222...
In base 5, 1/24 is 0.0101010101010101010101010101...
In base 6, 1/24 is 0.013.
Therefore base 6 is the lowest base in which 1/24 has a finite number of digits.
Today's number is the smallest base in which 1/10890 has a finite number of digits.
Note: 1/24 always represents 1 divided by twenty-four (ie the 24 is written in decimal).
Find a number base other than 10 in which 121 is a perfect square.
Let \(a_b\) denote \(a\) in base \(b\).
Find bases \(A\), \(B\) and \(C\) less than 10 such that \(12_A+34_B=56_C\).
Reverse bases again
Find three digits \(a\), \(b\) and \(c\) such that \(abc\) in base 10 is equal to \(cba\) in base 9?
Find two digits \(a\) and \(b\) such that \(ab\) in base 10 is equal to \(ba\) in base 4.
Find two digits \(c\) and \(d\) such that \(cd\) in base 10 is equal to \(dc\) in base 7.
Find two digits \(e\) and \(f\) such that \(ef\) in base 9 is equal to \(fe\) in base 5.