IMO 1975 · Problem 4

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.)

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{7} \]

Solution

Let \(N=4444^{4444}\), and let \(C\) be the sum of the digits of \(B\).

First, we bound \(C\). Since \[ N<(10^4)^{4444}=10^{17776}, \] the number \(N\) has at most \(17776\) digits. Each digit is at most \(9\), so \[ 1\le A\le9\cdot17776=159984<10^6. \] Thus \(A\) has at most six digits, giving \(1\le B\le54\). Consequently, the tens digit of \(B\) is at most \(5\) (taking it as zero when \(B<10\)), and its units digit is at most \(9\). Therefore \[ 1\le C\le14. \]

Every positive integer is congruent to its digit sum modulo \(9\): indeed, if its decimal digits are \(d_i\), then \[ \sum_i d_i10^i\equiv\sum_i d_i\pmod9. \] Applying this successively gives \[ C\equiv B\equiv A\equiv N\pmod9. \] Now \(4444\equiv7\pmod9\), \(7^3\equiv1\pmod9\), and \(4444=3\cdot1481+1\). Hence \[ N\equiv7^{4444} =(7^3)^{1481}\cdot7 \equiv7\pmod9. \] The only integer between \(1\) and \(14\) congruent to \(7\) modulo \(9\) is \(7\). Thus \(C=7\).

The solution was checked against John Scholes' IMO solution archive (Kalva).