Find the last three digits of the periods of the fractions $1 / 107, 1 / 131, 1 / 151$. (This can be done without calculating the preceding digits.)

#
The period of the fraction $1 / 107$ is obtained by dividing a number of the form 9...999 by 107. It is clear that the last digit is 7. By subtracting 749, we get a number of the form $9...925$, so the previous digit is 5. By subtracting 535, we get $9...939$, so the previous digit is 7.

We proceed similarly with the other two fractions.

## Answer

$757,229,649$.