IIT JEE , AIEEE Forum - Mathematics , Algebra

Anupamagrawal13a

In the decimal system of numeration the number of 6 digit numbers in which the sum of the digits is divisible by 5 is

Ans-180000

Please solve this question with good explaination.

Rates to all efforts

pannaguma

let the sum of the first five numbers be X. so the last digit can be filled in 10 ways. let it be Y.

assume X is divisible by 5. for X+Y to be divisible, only 0, or 5 can be the last number.

similarly if X not divisible by 5, X+Y will be divisible by only two nos out of 10 different last digit.

hence number of 6 digit numbers in which the sum of the digits is divisible by 5, is simply 1/5th the no of 6 digit nos.

which is 9*10^5 /5 = 180000.

Anupamagrawal13a

More explaination needed................

Please reply.

pannaguma

hey this question is more of a logic question.
for any 5 digits in the first five places, last digit can be put in ten ways of which only two will satisfy required condition.

eg, X=18, now Y can be only 2 or 7.
similar logic extends to all other numbers.

(**338**)

Here's the answer, rate me if you understand, or nudge me if u want an explanation...

pannaguma

12121212, please read the ques properly, it says SUM of the digits of the no.

NOT the number itself.

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yea thanks i made a mistake, but when i calculated orally (taking a diff question all together) and got the same answer 1,80,000 i hurriedly posted the answer.

but now i think the correct way to the question will be :

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No i guess its still incorrect....let me give another thought, i'll repost in a while...

(**338**)

the below method gives the answer, though it's very long...

i'll tell a shorter method once i find...

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