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`        the four digit no. 2652 is such that any two consecutive digits from it make a multipleof 13. another no. n has this same property, is 100 digits long and begins in a 9. the last digit of n is ? pls explain how this type of Qu are solved ? `
7 years ago

110 Points
```										such questions are solved by pure logic.there is no set formula for themnowwhich digit should come after 91?rightand after 13after 3?9after 9?1..so the series continuesthe number is9139139139.........upto hundred digits.the 99th digit would 3and hundredth? well it should be 9 right??simple!!CoolDifficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.DO APPROVE IF THE ANSWER HELPS YOU.Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.Regards,Askiitians ExpertsRahul- IIT Roorkee
```
7 years ago
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