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Missing Duodecimal Digit Determination (Posted on 2022-08-04) Difficulty: 3 of 5
If the duodecimal number A7?B5? is divisible by (83)base 12, then determine the missing digits.
** Each question mark represents a duodecimal digit, whether same or different.

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution solution | Comment 3 of 6 |
10*12^5 + 7*12^4 + x*12^3 + 11*12^2 + 5*12 + y is to be divisible by 8*12+3, which is 99.

Without the terms involving x and y:

10*12^5 + 7*12^4 + 11*12^2 + 5*12 = 2635116, which is congruent to 33 mod 99.

x*12^3 + y must be congruent to 66 mod 99. Of course, in addition each of x and y must be in the range 0 to 11.

12^3 = 1728 is congruent to 45 mod 99. If we try to leave well enough alone and let x=1 and make y congruent to 21 mod 99 we won't succeed in getting a value in the proper range.

Clearly x=0 won't work either as that would need y congruent to 66 mod 99.

We need to try every x to see what y is needed for that x.

Well what use is a computer if you don't use it for the grunt work. Without writing a script (program), just type:

>> for x= 1:11 y=mod(66-45*x,99); disp([x y])
end
     1    21
     2    75
     3    30
     4    84
     5    39
     6    93
     7    48
     8     3
     9    57
    10    12
    11    66
    
Clearly x=8 and y=3.

Verified with

>> mod(base2dec('a78b53',12),base2dec('83',12))
ans =
     0
     
BTW, when divided by 83 base 12 the answer is 13599 base 12.

In base 10 that's 2648943/99 = 26757
     

  Posted by Charlie on 2022-08-04 09:31:39
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