# 1994 AJHSME Problems/Problem 25

## Problem

Find the sum of the digits in the answer to

$\underbrace{9999\cdots 99}_{94\text{ nines}} \times \underbrace{4444\cdots 44}_{94\text{ fours}}$

where a string of $94$ nines is multiplied by a string of $94$ fours.

$\text{(A)}\ 846 \qquad \text{(B)}\ 855 \qquad \text{(C)}\ 945 \qquad \text{(D)}\ 954 \qquad \text{(E)}\ 1072$

## Solution 1

Notice that:

$9 \cdot 4 = 36$ and $3+6 = 9 = 9 \cdot 1$

$99 \cdot 44 = 4356$ and $4+5+3+6 = 18 = 9 \cdot 2$

So the sum of the digits of $x$ 9s times $x$ 4s is simply $x \cdot 9$ (Try to find the proof that it works for all values of $x$ ~MATHWIZARD10).

Therefore the answer is $94 \cdot 9 = \boxed{\text{(A)}\ 846.}$

## Solution 2

$$\underbrace{9999\cdots 99}_{94\text{ nines}} \cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = (10^{94}-1)\cdot \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{94\text{ fours}} \underbrace{000\cdots 0}_{94\text{ zeros}} - \underbrace{4444\cdots 44}_{94\text{ fours}} = \underbrace{444\cdots 4}_{93\text{ fours}} 3 \underbrace{555\cdots 5}_{93\text{ zeros}}6$$

$$4 \cdot 93 + 3 + 5 \cdot 93 + 6 = 9 \cdot 94 = \boxed{\text{(A)}\ 846}$$