All but One

Relevant wiki: Modular Arithmetic - Problem Solving - Basic

To find the missing digit, we use the fact that the digital sum of any integer $N$ is congruent to $N \! \pmod{9}.$ Here's a quick proof of why this is:

Let $N = 10^n a_n + 10^{n - 1}a_{n - 1} + \cdots + 10a_1 + a_0,$ where the $a_i$ are digits i.e. integers between 0 and 9 inclusive. Taking this expression modulo 9 yields

$\begin{aligned} N &\equiv 10^n a_n + 10^{n - 1}a_{n - 1} + \cdots + 10a_1 + a_0 \\ &\equiv 1^n a_n + 1^{n - 1}a_{n - 1} + \cdots + 1a_1 + a_0 \\ &\equiv a_n + a_{n - 1} + \cdots + a_1 + a_0 \! \! \! \! \pmod{9}. \end{aligned}$

Since $a_n + a_{n - 1} + \cdots + a_1 + a_0$ is precisely the digital sum of $N,$ we conclude that the digital sum of any positive integer is congruent to that integer modulo 9.

Now, because

$2^{29} \equiv 2^{27}\cdot2^2 \equiv (2^3)^9 \cdot 4 \equiv (-1)^9 \cdot 4 \equiv -4 \! \! \! \! \pmod{9},$

the digital sum of $2^{29}$ is equivalent to -4 modulo 9. It is given in the problem that $2^{29}$ contains all digits between 0 and 9 except for one. (Note that $10^8 < 2^{29} < 10^9,$ so $2^{29}$ has 9 digits, meaning every digit except for one of them must be included in it.) The sum of all these digits is congruent to 0 mod 9, so the missing digit $d$ is the one that will make $2^{29} + d$ be divisible by 9. This digit must be congruent to 4 modulo 9, and since it is a digit, the only possible value that works is $d = \boxed{4}.$

$2^{29} = 536870912$ .

Thus, the missing digit is $\boxed{4}$ .

Relevant wiki: Euler's Totient Function

Because $\phi(9) = 6$ we may compute $2^{29} \equiv 2^{-1} \equiv 5 \mod 9$ . On the other hand, a number mod 9 is equal to the sum of digits and the sum of the digits of this number will be $0 + 1 + ... + 9 - x = 45 - x$ where x is the digit that is missing. We can now solve $45 - x \equiv 5 \mod 9 \implies x \equiv 4 \mod 9 \implies x = 4$ .

First, we can expand $2^{29}$ as $2^9 \times 2^{10} \times 2^{10} = 512 \times 1024 \times 1024$ which is approximately 9-digit number.

So, as the question states, all of the distinct number appear except one number is missing.

So, do this: $2^{29} \equiv 4 \times 8^{9} \equiv 4 \times (-1)^9 \equiv 4 \times (-1) \equiv -4 (mod 9)$

There is a missing $\boxed {4}$

2^n Mod(9) has the progression 2,4,8,5,1 repeating Mod(5), 29 Mod(5) = 4, 4th member of progression is 5, 2^29 must then be 5 Mod(9) = -4 Mod(9), Therefore, missing digit is 4

As a bash one-liner:

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num=`dc <<< "2 29 ^ p"`; for i in `seq 0 9`; do echo $num | grep $i 1&>0 || break; done; echo "$i not found"

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4 not found

Use a calculator to find the decimal equivalent of 2^29. It's 536870912. So 4 is the missing digit.

To me it is not clear from the problem that these digits only appear once. Hence it is necessary to take $log_{10}$ , or to write: $2^{29} = \frac{(2^{10})^3}{2} = \frac{{1024}^3}{2}$ to deduce that the number in question has exactly 9 digits.

2^29=(2^7)^4 2=(128^4) 2=268435456*2=536870912 The only digit that does not appear in this number is 4.

I do web related development with html, javascript and php. So, it is trivial to execute alert(Math.pow(2,29)) in javascript in and read the answer to find out which digit is missing.

2^29 = 536870912 No. 4 is missing...

Since $2^3 \equiv -1$ mod 9, we have $2^{29} = (2^3)^9\cdot 2^2 \equiv (-1)^9\cdot 4 = -4\ \text{mod}\ 9.$ The sum of all digits $0,\dots,9$ is a multiple of 9, as is easily checked by paring them up ( $0 + 9, 1 + 8$ , etc.).

Therefore, the missing digit should be $0 - (-4) = 4$ mod 9.

All it requires is two multiplications. 2^29 = 2^10 * 2^9 * 2^10. I assume everyone knows that 2^10 = 1024 and 2^9 = 512. Two minutes with a biro and a scrap of paper. to solve this one.

Proper solution: by comparing $0 + \cdots + 9 = 45 \equiv 0 \pmod 3$ mod 3 and $2^{29} \equiv -1 \pmod 3$ the missing number should be $4 \equiv 1 \pmod 3$ .

Lazy solution: the sum from 0 to 9 is divisible by 3 where $2^{29}$ is not. If the answer were 2 or 8 then I wouldn't be able to pick one. So the answer is 4.

Of course if $\pmod 9$ was used instead of $\pmod 3$ the reasoning is invalid. But I was too lazy to even think about $2^{29} \equiv -1 \pmod 3$

I have no deep understanding of this, but my solution gives me the correct digit.

If we use following formula:

$\frac{1}{(1-x)^2}$ and use a number smaller than 1, lets say " $\frac{1}{2}$ " in our case.

Then we will get 4 as our solution, which is the wanted digit. If that isnt a coincidence, can somebody explain it properly or debunk it?