- rvc

Hom many non-negative integers less than 10,000 are there such that the sum of the digits of the number is divisible by 3?

- jamiebookeater

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- rvc

The options are
A. 1112
B. 2213
C. 2223
D. 3334

- math&ing001

Hint: If a number is divisible by 3, the sum of its digits are divisible by 3 and the other way round works too.

- rvc

i know that
but how will i find the sum?

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## More answers

- math&ing001

I was hoping you'd figure it out yourself. Solve 3k<=10,000 for k, with k a natural number.

- rvc

Ah
i have to solve 3k=10,000?

- math&ing001

Yep

- rvc

3333.3333

- rvc

but how u figured?

- math&ing001

Yeah that's why I put the inferior sign, round it to the inferior natural number.

- ikram002p

well sum of the digits divisible by 3 means the number itself divisible by 3

- mathmath333

does it includes 0

- rvc

@ikram002p please explain

- ikram002p

so
3,6,9,.......,9,999
all are divisible by 3 and there sum of digits is divisible by 3 as well

- rvc

yep

- math&ing001

Here's an example 18 is divisible by 3 : 1+8 = 9 is divisible by 3.

- ikram002p

:) so u can count them now ?

- mathmath333

read the divisiblity for 3 here https://en.wikipedia.org/wiki/Divisibility_rule

- ikram002p

here is a trick
3,6,9,.......,9,999 = 3*(1,2,3,......,1111)

- ikram002p

ps:- dont forget the zero ;)

- mathmath333

actually that equals 3333 from \(3,6,\cdots\ 9999\)

- ikram002p

oh sorry i made a typo :P

- ikram002p

so it would be 3333+(Zero count 1)=3334

- ikram002p

it says sum of digits divisible by 3 not equal 3
3|0

- rvc

:/

- rvc

ikky i get ur explanation
3(1,2...)
3k

- ikram002p

so now u can count them ?

- ikram002p

i made a typo first here
3,6,9,.......,9,999 = 3*(1,2,3,......,3333)

- ikram002p

and dont forget to add the zero :D

- rvc

yay
got it :*

- ikram002p

good :D

- rvc

Thank you so much to @ikram002p @Loser66 @mathmath333 @math&ing001 and that unknown user(i know who is it)

- ikram002p

you are the most welcome ;)

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