A number when successively divided by $9$, $11$ and $13$ leaves remainders $8$, $9$ and $8$ respectively A number when successively divided by $9$, $11$ and $13$ leaves remainders $8$, $9$ and
$8$ respectively.
The answer is $881$, but how? Any clue about how this is solved?

First when the number is divided by 9. the remainder is 8.
So N = 9x+8.
Similarly, next x = 11y+9, and y=13z+8.
So N = 99y+89 = 99(13z+8)+89 = 1287z+792+89 = 1287z+881.
So N is of the form, 1287*(A whole Number)+881.
If you need to find the minimum number, then it would be 881.
Hope that helps.
