Question

find the smallest 5 digit number which when divided by 13 leaves 4 as a remainder and when divisible by 17 leaves 5​

Answer 1

Step-by-step explanation:

I’ll assume you want the five digit integer closest to zero that fits your requirements.

Part 1

The smallest five digit number is 10000.

10000/11=909 and change.

The next integer after 909 is 910

910*11=10010

10010+5=10015

10015 is the smallest five digit number which, when divided by 11, leaves 5 as a remainder.

Part two

10015/13=770 and change

771 is the next integer after 770

771*13=10023

10023–10015=8

8 is the number which must be added to 10015 to make it evenly divisible by 13

Note

If you’re willing to accept adding a negative as fitting the requirement, you can also add -5 to 10015. This would get you 10010, which in addition to being evenly divisible by 11 is evenly divisible by 13, and is the smallest such 5 digit number.

Note two

Those two numbers, 8 and -5, are just the two closest to our original number. Actually any integer n will work in the formula 13n-5, there are infinite such numbers… because the question did not specify that our final number must result in a five digit number, much less that it must be the smallest such number.