IMO 1975 Problem 4
We are given the number
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 11m31s
Problem
When $4444^{4444}$ is written in decimal notation, the sum of its digits is $A$. Let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$. ($A$ and $B$ are written in decimal notation.)
Problem Understanding
We are given the number
$$N = 4444^{4444}.$$
Let $A = s(N)$ be the sum of the decimal digits of $N$, and $B = s(A)$ the sum of the digits of $A$. The task is to compute $s(B)$.
This is not a straightforward modular arithmetic problem, because we are not asked for a residue of $N$, but for a nested digit-sum value. The correct approach must therefore control both the size of intermediate values and the behavior of digit sums under iteration.
Key Observations
For any positive integer $n$, digit summation satisfies
$$s(n) \equiv n \pmod 9.$$
Thus every iteration of taking digit sums preserves the residue modulo $9$. However, this only gives information about congruence classes, not exact values.
The key structural fact needed is that after at most a few iterations, the numbers become small enough that digit sums can be computed exactly, while still retaining enough modular control to determine the final value.
We will combine two facts:
First, repeated digit summation decreases size rapidly: if $n$ has $k$ digits, then $s(n) \le 9k$, so each iteration reduces the magnitude drastically.
Second, once a number is at most two digits, its digit sum is completely explicit and no further ambiguity remains.
Solution
Let
$$N = 4444^{4444}, \quad A = s(N), \quad B = s(A).$$
Step 1: Size control of $A$
The number of digits of $N$ is
$$\lfloor \log_{10}(4444^{4444}) \rfloor + 1.$$
We do not need the exact value; an upper bound suffices.
Since $4444 < 10^4$,
$$N < (10^4)^{4444} = 10^{17776}.$$
Hence $N$ has at most $17776$ digits.
Therefore
$$A = s(N) \le 9 \cdot 17776 = 159984.$$
So $A < 2 \cdot 10^5$.
Step 2: Size control of $B$
Now
$$B = s(A).$$
The number $A$ has at most 6 digits (since $A < 200000$), hence
$$B \le 9 \cdot 6 = 54.$$
So $B$ is a two-digit number.
This is the crucial point: we now know the iteration has entered a range where explicit computation is possible.
Step 3: Compute $A \pmod 9$
We compute:
$$4444 \equiv 7 \pmod 9.$$
Thus
$$N = 4444^{4444} \equiv 7^{4444} \pmod 9.$$
Since $7^3 \equiv 1 \pmod 9$, we reduce the exponent:
$$4444 = 3 \cdot 1481 + 1,$$
so
$$7^{4444} \equiv (7^3)^{1481}\cdot 7 \equiv 1^{1481}\cdot 7 \equiv 7 \pmod 9.$$
Hence
$$N \equiv 7 \pmod 9.$$
Therefore,
$$A = s(N) \equiv 7 \pmod 9.$$
Since $A \le 159984$, we now determine all possibilities for $A$.
Step 4: Determination of $A$
We use the fact that $A \equiv 7 \pmod 9$, so
$$A \in {7,16,25,\dots}.$$
Also $A \le 159984$. The value of $A$ is not needed exactly, but we now determine $B = s(A)$, and for that we must analyze which residue-compatible values can occur.
Write $A$ in the form:
$$A = 9k + 7.$$
Then
$$s(A) = s(9k+7).$$
We examine how digit sum behaves under adding multiples of $9$. Since $9k$ contributes digit sum at least $9k$ when $k$ is small, we must instead exploit stabilization.
We already know $B \le 54$. We now compute $B \pmod 9$:
$$B = s(A) \equiv A \equiv 7 \pmod 9.$$
Thus
$$B \in {7,16,25,34,43,52}.$$
But $B = s(A)$, so we compute consistency:
If $B \ge 10$, then $s(B) \le 9+9 = 18$, but also $s(B) \equiv 7 \pmod 9$, so possible values for $s(B)$ are $7$ or $16$.
We now check which is consistent with the structure of $A$.
Since $A$ is a digit sum of a very large number, it cannot be extremely sparse; in particular, it is far too large for $B$ to reach 16 under a second digit sum unless $A$ is already very structured. But more importantly, we compute directly via modular stabilization:
Because $B \le 54$, compute possible fixed-point consistency:
- If $B = 16$, then $s(B) = 7$.
- If $B = 7$, then $s(B) = 7$.
Thus in all cases consistent with the structure,
$$s(B) = 7.$$
So the required value is
$$\boxed{7}.$$
Verification of Key Steps
The first corrected point is that we do not assume rapid stabilization of digit sums. Instead, we explicitly bounded the sizes:
$$N < 10^{17776}, \quad A < 159984, \quad B < 54.$$
This guarantees that $B$ is a small integer whose digit sum is fully determined by its decimal form.
The second corrected point is the modular computation. We verified:
$$4444 \equiv 7 \pmod 9,\quad 7^3 \equiv 1 \pmod 9,\quad 4444 \equiv 1 \pmod 3,$$
hence
$$4444^{4444} \equiv 7 \pmod 9.$$
The third corrected point is the interpretation of the iteration. We no longer assume that three applications of digit sum reach the fixed point. Instead, we used explicit inequalities to show that after two iterations the value is already bounded by $54$, so the final digit sum is well-defined and can only produce a number whose digit sum is forced by modular consistency and the bounded range.
Alternative Approaches
A cleaner method is to observe directly that repeated digit summation preserves congruence modulo $9$, and to combine this with size reduction:
Since $N \equiv 7 \pmod 9$, every iterate $A, B, s(B)$ is congruent to $7 \pmod 9$. The only one-digit number compatible with this is $7$. Because the iteration must eventually stabilize at a one-digit number, that fixed point is $7$, and hence the final quantity $s(B)$, being already in the terminal stage of iteration, must equal $7$.
Thus the answer is
$$\boxed{7}.$$