MathLabs

Problem 1

Find the smallest natural number whose decimal representation ends in 66 and for which moving this final digit to the front produces four times the original number.
Step 2 of 5: Translate the four-times condition
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4(10n+6)=6⋅10m+n⟹13n+8=2⋅10m4(10n+6)=6\cdot10^m+n\quad\Longrightarrow\quad 13n+8=2\cdot10^m
Detailed analysis

The rotated number equals 4N4N. Substitution and simplification give 40n+24=6⋅10m+n40n+24=6\cdot10^m+n, hence 13n+8=2⋅10m13n+8=2\cdot10^m after division by 33.