*Taking Three as the subject to reason about —
A convenient number to state —
We add Seven, and Ten, and then multiply out
By One Thousand diminished by Eight.*

*The result we proceed to divide, as you see,
By Nine Hundred and Ninety Two:
Then subtract Seventeen, and the answer must be
Exactly and perfectly true.*

Not all of Thomas Cranmer’s 42 Articles made it into the Elizabethian 39 Articles of the Anglicans.

19840_{ 10} = 231504_{ 6} .

231504_{ 10} can be devided by 42_{ 10} or by 39_{ 10} without remainder.

But did Lewis Carroll know that? 😉

(x+7+10)×(1000-8)/992-17 = x,

for x=3:

3 | 3 |

+ 7 | 10 |

+ 10 | 20 |

× (1000-8) | 19840 |

/ 992 | 20 |

– 17 | 3 |

100110110000000_{ 2} |
1111100000_{ 2} |

1000012211_{ 3} |
1100202_{ 3} |

10312000_{ 4} |
33200_{ 4} |

1113330_{ 5} |
12432_{ 5} |

231504_{ 6} |
4332_{ 6} |

111562_{ 7} |
2615_{ 7} |

46600_{ 8} |
1740_{ 8} |

30184_{ 9} |
1322_{ 9} |

19840_{ 10} |
992_{ 10} |

139A7_{ 11} |
822_{ 11} |

B594_{ 12} |
6A8_{ 12} |

9052_{ 13} |
5B4_{ 13} |

7332_{ 14} |
50C_{ 14} |

5D2A_{ 15} |
462_{ 15} |

4D80_{ 16} |
3E0_{ 16} |

40B1_{ 17} |
376_{ 17} |

3744_{ 18} |
312_{ 18} |

2GI4_{ 19} |
2E4_{ 19} |

29C0_{ 20} |
29C_{ 20} |

22KG_{ 21} |
255_{ 21} |

1ILI_{ 22} |
212_{ 22} |

1EBE_{ 23} |
1K3_{ 23} |

1AAG_{ 24} |
1H8_{ 24} |

16IF_{ 25} |
1EH_{ 25} |

1392_{ 26} |
1C4_{ 26} |

105M_{ 27} |
19K_{ 27} |

P8G_{ 28} |
17C_{ 28} |

NH4_{ 29} |
156_{ 29} |

M1A_{ 30} |
132_{ 30} |

KK0_{ 31} |
110_{ 31} |

JC0_{ 32} |
V0_{ 32} |

I77_{ 33} |
U2_{ 33} |

H5I_{ 34} |
T6_{ 34} |

G6U_{ 35} |
SC_{ 35} |

FB4_{ 36} |
RK_{ 36} |

2024-02-14