The Puzzle

The number \( 2^{29} \) is a 9-digit number where each digit appears exactly once, except one digit is missing.

Question: Without computing \( 2^{29} \) directly, which digit is missing?

Key Insight: Sum of Digits from 0–9

The sum of all digits from 0 through 9 is:

\[ 0 + 1 + 2 + \dots + 9 = \frac{9 \cdot 10}{2} = 45 \]

If \( 2^{29} \) contains 9 distinct digits, then one digit is missing from the complete set \( {0,1,\dots,9} \).

If we can compute the sum of the digits of \( 2^{29} \), then the missing digit is:

\[ 45 - \text{sum of digits of } 2^{29} \]

Step 1: Compute \( 2^{29} \mod 9 \)

The digital root or digit sum modulo 9 is congruent to the number modulo 9:

  • \( 2^3 = 8 \Rightarrow 2^6 = 64 \Rightarrow 2^9 = 512 \)
  • But easier: \( 2^6 = 64 \equiv 1 \mod 9 \), so the cycle repeats every 6

Now:

\[ 29 \equiv 5 \mod 6 \Rightarrow 2^{29} \equiv 2^5 = 32 \equiv 5 \mod 9 \]

So, the sum of the digits of \( 2^{29} \) must be congruent to 5 modulo 9.

Step 2: Find Which Digit to Remove

Let’s subtract each digit from 45 and check which result is \( \equiv 5 \mod 9 \):

  • \( 45 - 0 = 45 \equiv 0 \mod 9 \)
  • \( 45 - 1 = 44 \equiv 8 \mod 9 \)
  • \( 45 - 2 = 43 \equiv 7 \mod 9 \)
  • \( 45 - 3 = 42 \equiv 6 \mod 9 \)
  • \( 45 - 4 = 41 \equiv \boxed{5} \mod 9 \)
  • …

Only \( 45 - 4 = 41 \equiv 5 \mod 9 \)

Final Answer

\[ \boxed{4} \]

The missing digit in \( 2^{29} \) is 4.

Reference