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Question Number 118019 by prakash jain last updated on 14/Oct/20

How many n digit integers are divisible by 9?

$$\mathrm{How}\:\mathrm{many}\:{n}\:\mathrm{digit}\:\mathrm{integers}\:\mathrm{are} \\ $$$$\mathrm{divisible}\:\mathrm{by}\:\mathrm{9}? \\ $$

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

$${d}_{\mathrm{1}} {d}_{\mathrm{2}} ....{d}_{{n}} \:\mathrm{is}\:{n}\:\mathrm{digit}\:\mathrm{number}.\:\mathrm{where} \\ $$$${d}_{\mathrm{1}} ,{d}_{\mathrm{2}} ...,{d}_{{n}} \:\mathrm{are}\:\mathrm{digits}. \\ $$$$\mathrm{Express}\:\mathrm{result}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n}. \\ $$$$\mathrm{example}\:\mathrm{for}\:{n}=\mathrm{2} \\ $$$$\mathrm{18},\mathrm{27},\mathrm{36},\mathrm{45},\mathrm{54},\mathrm{63},\mathrm{72},\mathrm{81},\mathrm{90},\mathrm{99} \\ $$

Commented by mr W last updated on 14/Oct/20

9 has only one digit. 99 is valid.

$$\mathrm{9}\:{has}\:{only}\:{one}\:{digit}.\:\mathrm{99}\:{is}\:{valid}. \\ $$

Commented by prakash jain last updated on 14/Oct/20

Yes. my mistake.

$$\mathrm{Yes}.\:\mathrm{my}\:\mathrm{mistake}.\: \\ $$

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,...

$$\mathrm{It}\:\mathrm{is}\:\mathrm{infinite}\:\mathrm{many}\:\mathrm{because}\:\mathrm{the}\:\mathrm{condition} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{number}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{9}\:\mathrm{that}\:\mathrm{is} \\ $$$$\mathrm{the}\:\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its}\:\mathrm{digits}\:\:\mathrm{divisible}\:\mathrm{by9} \\ $$$$\mathrm{Example}:\mathrm{9},\mathrm{18},\mathrm{108},\mathrm{1008},\mathrm{10008},... \\ $$

Commented by prakash jain last updated on 14/Oct/20

Sorry about not being clear on question. Added clarification on question

$$\mathrm{Sorry}\:\mathrm{about}\:\mathrm{not}\:\mathrm{being}\:\mathrm{clear}\:\mathrm{on}\:\mathrm{question}. \\ $$$$\mathrm{Added}\:\mathrm{clarification}\:\mathrm{on}\:\mathrm{question} \\ $$

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

$${it}'{s}\:{easier}\:{than}\:{one}\:{may}\:{think}. \\ $$$${the}\:{first}\:{one}\:{is}\:\mathrm{10}^{{n}−\mathrm{1}} +\mathrm{8} \\ $$$${the}\:{second}\:{one}\:{is}\:\mathrm{10}^{{n}−\mathrm{1}} +\mathrm{17} \\ $$$$... \\ $$$${the}\:{last}\:{one}\:{is}\:\mathrm{10}^{{n}} −\mathrm{1} \\ $$$$ \\ $$$${totally} \\ $$$$\frac{\left(\mathrm{10}^{{n}} −\mathrm{1}\right)−\left(\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{1}\right)}{\mathrm{9}}=\mathrm{10}^{{n}−\mathrm{1}} \:{numbers} \\ $$

Commented by prakash jain last updated on 14/Oct/20

Thanks.

$$\mathrm{Thanks}. \\ $$

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