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 5 of 5: Compute and verify the smallest number
n′=105−413=7692,N=10(2n′)+6=153846n'=\frac{10^5-4}{13}=7692,\qquad N=10(2n')+6=153846
Detailed analysis

For m=5m=5, n′=7692n'=7692, so n=15384n=15384 and N=153846N=153846. Moving the final 66 gives 615384=4⋅153846615384=4\cdot153846, and no smaller digit length was possible; hence this is the smallest number.