Number Theory I

Let the number be of the form $13k+3$ and $11j+7$ . This implies $13k=11j+4$ . Note that the first number of this sort is $26$ . Hence all the other numbers in this series are $143l + 26$ ; since the LCM is $143$ . Don't forget the 3 in the original number; this gives us a number of the form $143l + 29$ . Now, again apply the same modular arithmetic rule along with $7a+5$ to get $\boxed{887}$ . Please note that this LCM method would work for any co-prime numbers. So the next number satisfying the given set of conditions would be $887 + 143 \times 7$

For mod 11, the numbers would be 18,29,40.....
For mod 13, the numbers would be 16,29,.............. <..>29 satisfies 11 and 13. But
29 = 1 (mod 7 ). <.....> Next satisfactory number will be 11 * 13 = 143 more.. 143= 3 (mod 7)
For every addition of 143 to 29, we add 3 (mod 7). <..>For 29 we have 1 (mod 7 )
To have 5 (mod 7) we need .....12, 19, 26, 33, 40, ..... which should be 1 (mod 3) ..<..>1 for 29 and 3 for every 143.
The first to give this is 19 = 1 (mod 3) and.. 19 - 1 = 18 <..> 18/3 = 6. So, 6 of 143. <..> 29 + 6*143 = 887.