Three adjacent digits of a five digit number are 4s. find the five digit number if it is divisible by 315.

First start with number 315. You can see that 315=3·3·5·7. If the five digit number is divisible by 315, then it must be divisible by 3, 9, 5 and 7.

1. When number is divisible by 5, then the number's last digit is either 0 or 5. 

2. When number is divisible by 3, the sum of all the individual digits is evenly divisible by 3.

3. When number is divisible by 9, the sum of all the individual digits is evenly divisible by 9.

Since three adjacent digits of a five digit number are 4s, then yhis number could be 444*0, 444*5, *4440, *4445 (note that **444 couldn't be, because then this number has last digit 4 instead of 0 or 5).

For the number 444*0 find all possibilities for *: 4+4+4+0+(*)=12+(*) should be divisible by 9, then *=6 and you get 44460, but this number isn't divisible by 7.

For the number 444*5 find all possibilities for *: 4+4+4+5+(*)=17+(*) should be divisible by 9, then *=1 and you get 44415 and 44415÷7=6345. Also you can see that 44415÷315=141. This number is suitable.

For the number *4440 find all possibilities for *: 4+4+4+0+(*)=12+(*) should be divisible by 9, then *=6 and you get 64440, but this number isn't divisible by 7.

For the number *4445 find all possibilities for *: 4+4+4+5+(*)=17+(*) should be divisible by 9, then *=1 and you get 14445, but this number isn't divisible by 7.

44415

Iam pretty sure its a sorry if im wrong