## ParthKohli 4 years ago I just discovered something which I am unable to prove. Any natural number $$n$$ can be represented as a multiple of 9 + Sum of the digits of $$n$$.

1. ParthKohli

For example: 9 = 0 + 9 40 = 36 + 4 + 0 120 = 117 + 2 + 0

2. anonymous

I got you but 120 = 117 + 1 + 2 + 0

3. ParthKohli

Oops.

4. anonymous

45 = 45 + 4 + 5 ???

5. ParthKohli

No, 45 = 36 + 4 + 5

6. anonymous

Kidding: 45 = 36 + 4 + 5

7. ParthKohli

Do you use induction to prove?

8. anonymous

what about the numbers below 9?

9. ParthKohli

8 = 0 + 8

10. anonymous

3 = 9(0) + 3

11. ParthKohli

0 is a multiple of 9. 9 * 0 = 0

12. anonymous

isnt that cheating lol? everything is a multiple of zero then

13. ParthKohli

But I need to prove this thing. How do you do that?

14. anonymous

98 = 90 + 9 + 8 ???

15. anonymous

that's 107?

16. anonymous

more like 81 + 9 + 8

17. anonymous

98 = 81 + 9 + 8

18. ParthKohli

How do you prove it?

19. UnkleRhaukus

$\text{decimal}$

20. Mimi_x3

Can you use Mathematical Induction to prove that? lol

21. ParthKohli

I have no idea.

22. anonymous

He is saying natural number @UnkleRhaukus

23. Mimi_x3

I don't think so Parth; you can't prove that with MI, not certain though.

24. anonymous

I guess, Mukushla will give something to cheer upon..

25. anonymous

i have a feeling it has something to do with mods

26. UnkleRhaukus

our number system is to base 10

27. anonymous

sorry i lost my connection if n is a m+1 digit number $$n=(a_{m}a_{m-1}....a_{1}a_{0})=10^m a_{m}+10^{m-1} a_{m-1}+...+10a_{1}+a_{0}$$ now $$n-(a_{m}+a_{m-1}+...+a_{1}+a_{0})=(10^m -1)a_{m}+(10^{m-1}-1)a_{m-1}+...+9a_{1}=9k$$

28. anonymous

120=117+1+2+0

29. ParthKohli

I got a proof, finally. $$\color{Black}{\Rightarrow \bar{abc} = 100a + 10a + c }$$ $$\color{Black}{\Rightarrow 99a + a + 9b + b + c}$$ $$\color{Black}{\Rightarrow 9(11a + b) + (a + b + c)}$$ 11a + b is a multiple of 9.

30. ParthKohli

Is my proof good enough?

31. ParthKohli

10b*

32. anonymous

@ParthKohli u did my work for m=2

33. anonymous

$$(10^k - 1)$$ will be multiple of 9 always..

34. anonymous

@waterineyes thats right

35. anonymous

My guess is right: Mukushla has given something to cheer upon..

36. ParthKohli

Yeah.