+26367 10,11,12,13,14,15,16,17,18,19,20,21,?,31
Sum of digits is strictly greater than product of digits.
\(10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, \mathbf{30}, 31, \\ 40, 41, 50, 51, 60, 61, 70, 71, 80, 81, 90, 91, 100, 101, \\ 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, \\ 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 130, 131, \\ 140, 141, 150, 151, 160, 161, 170, \ldots\)
Example:
\(\begin{array}{|r|r|c|r|} \hline & \text{ Sum of digits } && \text{product of digits} \\ \hline 10 & 1+0 = 1 && 1\cdot 0 = 0 \\ & & 1 > 0 & \\\\ 11 & 1+1 = 2 && 1\cdot 1 = 1 \\ & & 2 > 1 & \\\\ 12 & 1+2 = 3 && 1\cdot 2 = 2 \\ & & 3 > 2 & \\ \ldots & && \\ \\ 20 & 2+0 = 2 && 2\cdot 0 = 0 \\ & & 2 > 0 & \\\\ 21 & 2+1 = 3 && 2\cdot 1 = 2 \\ & & 3 > 2 & \\ \ldots & && \\ \\ 30 & 3+0 = 3 && 3\cdot 0 = 0 \\ & & 3 > 0 & \\\\ 31 & 3+1 = 4 && 3\cdot 1 = 3 \\ & & 4 > 3 & \\ \hline \end{array}\)
+8 Your solution to answer is great . However i have other solution to this problem :
10+1^0 =11 , 11+1^1=12 , 12 +1^2 =13 , 13+1^3=14 .... we also have 20+2^0=21 ,21+2^1=23, 23+2^3=31
So my answer is 23 . Hope to see your response soon ! Thanks
