Divisibility by 15
Proof:
Let the number be abcde.
Remainderabcde can be written as a×10000 + b×1000 + c×100 + d×10 + e which can also be written as a×(15×666+10) + b×(15×66+10) + c×(15×6+10) + d×10 + e Every power of 10 i.e. 10^{x} when divided by 15 gives 10 as remainder,x>1 and is a positive integer.
Proof for remainder of 10^{x} when divided by 15
100 = 6×15 + 10 1000 = 10×(100) =10(15×6+10) =10×15×6 + 100 remainder = 10 10000 = 10×(1000) =10(15×6+10 + 100) =10×15×6×10 + 1000 remainder = 10, as 10 is remainder of 1000. Remainder when divided by 15 of 100 is 10, similarly of 1000 (other product has factor 15) the process continues as we get 100 by this process each time and 100 has remainder 10. Every product is divisible except a×10 + b×10 + c×10 + d×10 + e So the number is divisible if the sum is divisible by 15. The number is divisible by 15 if [(sum of all digits except the last digit)×10 + digit at unit place ] is divisible by 15.
Examples
