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How-many-n-digit-integers-are-divisible-by-9-




Question Number 118019 by prakash jain last updated on 14/Oct/20
How many n digit integers are  divisible by 9?
Howmanyndigitintegersaredivisibleby9?
Commented by prakash jain last updated on 14/Oct/20
d_1 d_2 ....d_n  is n digit number. where  d_1 ,d_2 ...,d_n  are digits.  Express result in terms of n.  example for n=2  18,27,36,45,54,63,72,81,90,99
d1d2.dnisndigitnumber.whered1,d2,dnaredigits.Expressresultintermsofn.exampleforn=218,27,36,45,54,63,72,81,90,99
Commented by mr W last updated on 14/Oct/20
9 has only one digit. 99 is valid.
9hasonlyonedigit.99isvalid.
Commented by prakash jain last updated on 14/Oct/20
Yes. my mistake.
Yes.mymistake.
Answered by 1549442205PVT last updated on 14/Oct/20
It is infinite many because the condition  for a number divisible by 9 that is  the  sum of its digits  divisible by9  Example:9,18,108,1008,10008,...
Itisinfinitemanybecausetheconditionforanumberdivisibleby9thatisthesumofitsdigitsdivisibleby9Example:9,18,108,1008,10008,
Commented by prakash jain last updated on 14/Oct/20
Sorry about not being clear on question.  Added clarification on question
Sorryaboutnotbeingclearonquestion.Addedclarificationonquestion
Answered by mr W last updated on 14/Oct/20
it′s easier than one may think.  the first one is 10^(n−1) +8  the second one is 10^(n−1) +17  ...  the last one is 10^n −1    totally  (((10^n −1)−(10^(n−1) −1))/9)=10^(n−1)  numbers
itseasierthanonemaythink.thefirstoneis10n1+8thesecondoneis10n1+17thelastoneis10n1totally(10n1)(10n11)9=10n1numbers
Commented by prakash jain last updated on 14/Oct/20
Thanks.
Thanks.

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