Question

A number being successively divided by 9 11 and 13 leaves 8 9 and 8 reminders respectively of the order of divisors is reversed then remainder is

Answer 1

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.

Answer 2

The new remainder would be 7,2,6

1) Let the number be N.

After first division N = 9x + 8

remaining number is x

After second division x = 11y+9

After third division y = 13z + 8

Hence N = 9(11y+9) + 8 = 99y + 89 = 99(13z +8) + 89 = 1287z  + 891

2) Reversing the order of divisors

After first division (1287z +891 )/13 = 99z + 68  with remainder 7

After second division (99z + 68)/11 = 9z + 6  with remainder 2

after third division (9z +6)/9 = z with remainder 6

3) Hence the new remainder would be 7,2,6