A7?B5? is divisible by the duodecimal number 83, which is equal to 8*12+3=99 in base 10.
Now, 99 =9*11, where gcd(9,11) = 1
In duodecimal parlance this translates to:
(83)base 12 = 9*B
Accordingly the given number must be separately divisible by each of 9 and B.
Now any exponent higher than 12 is divisible by 9, so that the number itself is divisible by 9 only when the number formed by the last two digits is divisible by 9.
Accordingly, 9 must divide the duodecimal number 5? which is 60+n in base 10 (replacing ? by n). It can easily be seen that 63 is divisible by 9, so that n=3.
So, the number is now A7?B53, which is divisible by B(11 in base 10). Since B is precisely 1 less than duodecimal number 10, we can use the method of "casting out B's" which is similar to casting out 9s in base ten.
Thus, replacing the question mark by m we have:
A+7+m+5+3 or, 21+m in duodecimal system. This must be evenly divisible by B
Since this duodecimal multiplication:
B*3 = 29 is true - we merely subtract the base twelve number 21 from the duodecimal number 29, to obtain m=8.
Consequently, A78B53 is our desired duodecimal integer, and 8 and 3 are the required missing digits (in order)
As a CHECK, we note that:
The duodecimal number A78B53 is equal to the base ten number 2648943, and 83 in base 12 is 99 in base ten.
Now, 2648943/99 = 26957 with no remainder.
Accordingly, the deduction of the missing values is accurate.
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For an alternative analytical methodology, refer to the solution submitted by xdog in this location.
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