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?
The options are A. 1112 B. 2213 C. 2223 D. 3334
Hint: If a number is divisible by 3, the sum of its digits are divisible by 3 and the other way round works too.
i know that but how will i find the sum?
I was hoping you'd figure it out yourself. Solve 3k<=10,000 for k, with k a natural number.
Ah i have to solve 3k=10,000?
Yep
3333.3333
but how u figured?
Yeah that's why I put the inferior sign, round it to the inferior natural number.
well sum of the digits divisible by 3 means the number itself divisible by 3
does it includes 0
@ikram002p please explain
so 3,6,9,.......,9,999 all are divisible by 3 and there sum of digits is divisible by 3 as well
yep
Here's an example 18 is divisible by 3 : 1+8 = 9 is divisible by 3.
:) so u can count them now ?
read the divisiblity for 3 here https://en.wikipedia.org/wiki/Divisibility_rule
here is a trick 3,6,9,.......,9,999 = 3*(1,2,3,......,1111)
ps:- dont forget the zero ;)
actually that equals 3333 from \(3,6,\cdots\ 9999\)
oh sorry i made a typo :P
so it would be 3333+(Zero count 1)=3334
it says sum of digits divisible by 3 not equal 3 3|0
:/
ikky i get ur explanation 3(1,2...) 3k
so now u can count them ?
i made a typo first here 3,6,9,.......,9,999 = 3*(1,2,3,......,3333)
and dont forget to add the zero :D
yay got it :*
good :D
Thank you so much to @ikram002p @Loser66 @mathmath333 @math&ing001 and that unknown user(i know who is it)
you are the most welcome ;)
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