Question

Bazil wrote two numbers in his notebook, {2}^{12}{3}^{2}{5}^{7}{7}^{5} and {2}^{3}{3}^{12}{5}^{2}{7}^{2}. After that he proceeded writing numbers in the notebook according to the following rule. Each time he can write down a positive number equal to the difference of any two numbers already written in the copybook. It is not allowed to repeat the numbers in the notebook. Find the sum of two smallest numbers that can be obtained in the notebook.

Answer 1

Given numbers: ${2}^{12}{3}^{2}{5}^{7}{7}^{5}\ and\ {2}^{3}{3}^{12}{5}^{2}{7}^{2}$

${2}^{12}{3}^{2}{5}^{7}{7}^{5}=48404160000000 \\{2}^{3}{3}^{12}{5}^{2}{7}^{2}=5208121800$

Next positive number$={2}^{12}{3}^{2}{5}^{7}{7}^{5}-{2}^{3}{3}^{12}{5}^{2}{7}^{2}$

Next positive number$={2}^{12}{3}^{2}{5}^{7}{7}^{5}-{2}^{3}{3}^{12}{5}^{2}{7}^{2}-{2}^{3}{3}^{12}{5}^{2}{7}^{2}$

$={2}^{12}{3}^{2}{5}^{7}{7}^{5}-{2}^{4}{3}^{12}{5}^{2}{7}^{2}$

And so on, we will have two positive numbers $={2}^{12}{3}^{2}{5}^{7}{7}^{5}-{2}^{11}{3}^{12}{5}^{2}{7}^{2}$ &

${2}^{12}{3}^{2}{5}^{7}{7}^{5}-{2}^{12}{3}^{12}{5}^{2}{7}^{2}$

Their sum$={2}^{12}{3}^{2}{5}^{7}{7}^{5}-{2}^{12}{3}^{12}{5}^{2}{7}^{2}+{2}^{12}{3}^{2}{5}^{7}{7}^{5}-{2}^{11}{3}^{12}{5}^{2}{7}^{2}$

$={2}^{13}{3}^{2}{5}^{7}{7}^{5}-{2}^{11}{3}^{12}{5}^{2}{7}^{2}(2+1) \\={2}^{13}{3}^{2}{5}^{7}{7}^{5}-{2}^{11}{3}^{13}{5}^{2}{7}^{2}$