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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?
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
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 :*
Thank you so much to @ikram002p @Loser66 @mathmath333 @math&ing001 and that unknown user(i know who is it)
you are the most welcome ;)