# A Missing Digit

### Problem

### Solution 1

We'll employ digital roots. Digital root $D(n)$ of an integer $n$ is, by definition, the sum of its digits computed recursively until only one digit remains. Importantly, $D(n)\equiv n\text{ mod } 9.$

Now, for the powers of $2$ we have the following digital roots:

$\begin{array}{cccccccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & \cdots & 29\\ \hline 1 & 2 & 4 & 8 & 7 & 5 & 1 & 2 & 4 & 8 & 7 & 5 & \cdots & 5 \end{array}$

This tells us $D\left(2^{29}\right)=5,$ $D\left(2^{29}\right)\equiv 5\text{ mod }9.$ Next observe that the sum of all ten decimal digits is $45,$ with $D(45)=9\equiv 0\text{ mod }9.$ From this we deduce that, if $d$ is the missing digit, then $D(2^{29})+d\equiv 0\text{ mod }9.$ Since, as we found, $D\left(2^{29}\right)=5,$ $d=4.$

Now you can check that $2^{29}=536,870,912.$

### Solution 2

$2^5=32\equiv 5 (mod~9);$ $2^6=64\equiv 1 (mod~9)$. Thus,

$2^{29}=(2^6)^4\cdot 2^5\equiv 5 (mod~9)\equiv -4(mod~9)$.

A number and the sum of all its digits are congruent to each other modulo $9$. Moreover, the sum of all digits is $45\equiv 0(mod~9)$. Thus, the unique digit that has to be missing to get us to $-4 (mod~9)$ from $0~(mod~9)$ is $4$.

### Acknowledgment

This is a problem from P. Winkler's Mathematical Mind-Benders (A K Peters/CRC Press, 2007). Solution 2 is by Amit Itagi.

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