Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
ParthKohli
Group Title
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 \(n\).
 2 years ago
 2 years ago
ParthKohli Group Title
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 \(n\).
 2 years ago
 2 years ago

This Question is Closed

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
For example: 9 = 0 + 9 40 = 36 + 4 + 0 120 = 117 + 2 + 0
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
I got you but 120 = 117 + 1 + 2 + 0
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
45 = 45 + 4 + 5 ???
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
No, 45 = 36 + 4 + 5
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Kidding: 45 = 36 + 4 + 5
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Do you use induction to prove?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
what about the numbers below 9?
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
8 = 0 + 8
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
3 = 9(0) + 3
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
0 is a multiple of 9. 9 * 0 = 0
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
isnt that cheating lol? everything is a multiple of zero then
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
But I need to prove this thing. How do you do that?
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
98 = 90 + 9 + 8 ???
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
that's 107?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
more like 81 + 9 + 8
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
98 = 81 + 9 + 8
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
How do you prove it?
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\text{decimal} \]
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
Can you use Mathematical Induction to prove that? lol
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I have no idea.
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
He is saying natural number @UnkleRhaukus
 2 years ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.1
I don't think so Parth; you can't prove that with MI, not certain though.
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
I guess, Mukushla will give something to cheer upon..
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i have a feeling it has something to do with mods
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
our number system is to base 10
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
sorry i lost my connection if n is a m+1 digit number \( n=(a_{m}a_{m1}....a_{1}a_{0})=10^m a_{m}+10^{m1} a_{m1}+...+10a_{1}+a_{0} \) now \( n(a_{m}+a_{m1}+...+a_{1}+a_{0})=(10^m 1)a_{m}+(10^{m1}1)a_{m1}+...+9a_{1}=9k \)
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
120=117+1+2+0
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Is my proof good enough?
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
@ParthKohli u did my work for m=2
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
\((10^k  1)\) will be multiple of 9 always..
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
@waterineyes thats right
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
My guess is right: Mukushla has given something to cheer upon..
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.