An algebra problem by أحمد الحلاق

$\begin{array} {crrrc} n & n \text{-digit of 3} & (n+1) \text{-digit of 6} & \text{Product }NM & \text{Sum of digits }S_d(n) \\ \hline 1 & 3 & 66 & 198 & 18 = 9(1+1) \\ 2 & 33 & 666 & 21978 & 27 = 9(2+1) \\ 3 & 333 & 6666 & 2219778 & 36 = 9(3+1) \\ 4 & 3333 & 66666 & 222197778 & 45 = 9(4+1) \\ 5 & 33333 & 666666 & 22221977778 & 54 = 9(5+1) \end{array}$

Consider the products $NM$ of first few $n$ -digit of 3 and $(n+1)$ -digit of 6 we get that the sum of digits of $NM$ , $S_d (n) = 9(n+1)$ . therefore, $S_d (61) = 9(62) = \boxed{558}$ .

But the claim that $S_d(n) = 9(n+1)$ for all $n \ge 1$ should be proven.

Let's define $X_k = \underbrace{3\ldots3}_{\text{k times}}$ and $Y_k = \underbrace{6\ldots6}_{\text{k+1 times}}$ , where $k$ is a positive integer.

Let's now manipulate their product.

$\begin{aligned}X_kY_k &= \underbrace{3\ldots3}_{\text{k times}} \times \underbrace{6\ldots6}_{\text{k+1 times}} \\ &= \underbrace{2\ldots2}_{\text{k times}} \times \underbrace{9\ldots9}_{\text{k+1 times}} \\ &= \underbrace{2\ldots2}_{\text{k times}} \times 10^{k+1} - \underbrace{2\ldots2}_{\text{k times}} \\ &= \underbrace{2\ldots2}_{\text{k times}} \times 10^{k+1} - 10^{k} +\underbrace{7\ldots7}_{\text{k times}} + 1 \\ &= \underbrace{2\ldots2}_{\text{k-1 times}} \times 10^{k+2} + 1 \times 10^{k+1} + 9 \times 10^{k} + \underbrace{7\ldots7}_{\text{k-1 times}} \times 10 + 8 \end{aligned}$

Hence, the digit sum of $X_kY_k$ is equal to $\left(k-1\right)\times2 + 1 + 9 + \left(k-1\right)\times7+8 = \left(k+1\right)\times9$ .

Since $N = X_{61}$ and $M = Y_{61}$ , the digit sum of $NM$ equals $62 \times 9 = \boxed{558}$ .