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 5 of 6: Compute the residue of

4444^{4444}

modulo 9

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4444 \equiv 7,\ 7^3\equiv 1 \pmod 9,\ 4444=3\cdot1481+1 \implies 4444^{4444}\equiv 7 \pmod 9

Detailed analysis

Since $4444 = 9 \cdot 493 + 7$ , $4444 \equiv 7 \pmod 9$ . Powers of $7$ modulo $9$ cycle with period $3$ : $7^1 \equiv 7$ , $7^2 \equiv 4$ , $7^3 \equiv 1 \pmod 9$ . Writing $4444 = 3 \cdot 1481 + 1$ , we get $4444^{4444} \equiv 7^{4444} = \left(7^3\right)^{1481} \cdot 7 \equiv 1^{1481} \cdot 7 = 7 \pmod 9$ .