Question

How many numbers lie between 100 and 10000 which when successively divided by 7, 11 and 13 leaves the respective remainders of 5, 6 and 7? ___

Answer 1

The least possible number can be obtained as

(((7 $\times$ 11) + 6) 7 + 5) = ((77 + 6) 7 + 5)
= (83 $\times$ 7 + 5) = (581 + 5) = 586
The general form for the higher numbers is
(7 $\times$ 11 $\times$ 13)m + 586 = (1001)m + 586
So, the numbers can be obtained by considering m = 0, 1, 2, 3, ... so the first number is 586 and the last number is 9595 which can be attained at m = 9. So there are total 10 such numbers lying between 100 and 10000.