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.)

Step 6 of 6: Combine the bound and the residue to pin down

S(B)

Previous step Next step

S(B)\le 12,\ S(B)\equiv 7\pmod 9,\ S(B)\ge 1 \implies S(B)=7

Detailed analysis

From Step 4, $S(B) \equiv 4444^{4444} \equiv 7 \pmod 9$ (Step 5). From Steps 2–3, $1 \le S(B) \le 12$ (it is a digit sum of a positive integer, so at least $1$ ). The only integer in $\{1, \ldots, 12\}$ congruent to $7$ modulo $9$ is $7$ itself (the next candidate, $16$ , already exceeds the bound). Therefore $S(B) = 7$ . $\blacksquare$