.
To solve the problem, use the divisibility by 9 rule in this extended form:
The remainder of division the number by 9 is the same as the remainder of the sum of its digits divided by 9.
It is very widely known rule. See, for example, the lesson Divisibility by 9 rule in this site.
The number 3851902343886132 (first 16-digit factor) has the sum of digits 66.
Hence, according to the divisibility rule above, its remainder of division by 9 is the same
as the remainder of the number 66 divided by 9, i.e. 66-7*9 = 66-63 = 3.
The number 5221791683705111 (second 16-digit factor) has the sum of digits 59.
Hence, according to the divisibility rule above, its remainder of division by 9 is the same
as the remainder of the number 59 divided by 9, i.e. 59-6*9 = 59-54 = 5.
It implies that the remainder of the product of these two 16-digit numbers when divided by 9
is 3*5 (mod 9) = 15 (mod 9), i.e. 6.
The sum of digits of the 32-digit product 20 113 831 625 748 808 690 240 050 420 652
( ! I temporarily inserted "0" (zero) at the missed position for a moment)
is 112, which gives the remainder 4 when divided by 9.
We need to get the remainder 6, so I need to replace the inserted "0" by "2" to get it.
Hence, the missed digit under the question is "2".
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