【Question 12】
It is known that $2^{29}$ is a nine-digit number consisting of nine different digits. Which digit is missing?
【Analysis and Solution】
Number Theory, Congruence.
$$
2^{1} \equiv 2(\bmod 9), \quad 2^{2} \equiv 4(\bmod 9), \quad 2^{3} \equiv 8(\bmod 9), \quad 2^{4} \equiv 7(\bmod 9), \quad 2^{5} \equiv 5(\bmod 9), \quad 2^{6} \equiv 1(\bmod 9) ;
$$
The remainder of $2^{n} \div 9$ cycles through “ $2,4,8,7,5,1$ ”;
Since $29 \div 6=4 \cdots \cdots \cdot 5$;
Therefore, the remainder of $2^{29} \div 9$ is the same as the remainder of $2^{5} \div 9$, which is 5;
If no digit is missing, the sum of all digits is $0+1+2+3+4+5+6+7+8+9=45$;
And $45 \div 9$ leaves a remainder of 0;
Thus, the missing digit in $2^{29}$ is $9-5=4$.
Of course, this can also be verified by calculation, $2^{29}=536870912$.