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Mathematics 8 Online
rvc (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?

rvc (rvc):

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

OpenStudy (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 (rvc):

i know that but how will i find the sum?

OpenStudy (math&ing001):

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

rvc (rvc):

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

OpenStudy (math&ing001):

Yep

rvc (rvc):

3333.3333

rvc (rvc):

but how u figured?

OpenStudy (math&ing001):

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

OpenStudy (ikram002p):

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

OpenStudy (mathmath333):

does it includes 0

rvc (rvc):

@ikram002p please explain

OpenStudy (ikram002p):

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

rvc (rvc):

yep

OpenStudy (math&ing001):

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

OpenStudy (ikram002p):

:) so u can count them now ?

OpenStudy (mathmath333):

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

OpenStudy (ikram002p):

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

OpenStudy (ikram002p):

ps:- dont forget the zero ;)

OpenStudy (mathmath333):

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

OpenStudy (ikram002p):

oh sorry i made a typo :P

OpenStudy (ikram002p):

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

OpenStudy (ikram002p):

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

rvc (rvc):

:/

rvc (rvc):

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

OpenStudy (ikram002p):

so now u can count them ?

OpenStudy (ikram002p):

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

OpenStudy (ikram002p):

and dont forget to add the zero :D

rvc (rvc):

yay got it :*

OpenStudy (ikram002p):

good :D

rvc (rvc):

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

OpenStudy (ikram002p):

you are the most welcome ;)

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