Let missing digits equal X and Y so base-12 A7?B5? becomes base-10
10*12^5 + 7*12^4 + X*12^3 + 11*12^2 + 5*12 + Y.
The divisor is base-12 83 which is 99=9*11 base-10.
To be divisible by 9 requires 60 + Y divisible by 9 so Y=3.
B and A are respectively 11 and 10 in base ten. Therefore, to be divisible by 11 requires 10+7+X+63 divisible by 11 so x=8.
The duodecimal number is then A78B53 which equals decimal 2648943=99*26757
***edited per request by puzzle author***
Edited on August 5, 2022, 12:50 pm
|
Posted by xdog
on 2022-08-04 09:08:51 |