InterviewSolution
Saved Bookmarks
| 1. |
Consider all `10` digit numbers formed by using all the digits 0, 1, 2, 3,…, 9 without repetition such that they are divisible by 11111, thenA. the digit in tens place for smallest number is 6B. the digit in tens place for largest number is 3C. total numbers of such numbers is 3456D. total numbers of such numbers is 365 |
|
Answer» Correct Answer - A::B::C Let `N=x_(1)x_(2)x_(3)………..x_(10)` be one of such number `sumx_(i)=45`, hence `N` is divisible by 9. Hence `N` should by divisible by `11111xx9=99999` Now `x_(1)x_(2)x_(3)………x_(10)=x_(1)x_(2)x_(3)x_(4)x_(5)xx10^(5)+x_(6)x_(7)x_(8)x_(9)x_(10)` `=x_(1)x_(2)x_(3)x_(4)x_(5)lt99999` `x_(6)x_(7)x_(8)x_(9)x_(10)lt 99999` `x_(1)x_(2)x_(3)x_(4)x_(5)+x_(6)x_(7)x_(8)x_(9)x_(10)lt 2xx99999` `x_(1)x_(2)x_(3)x_(4)x_(5)+x_(6)x_(7)x_(8)x_(9)x_(10)=99999` `x_(1)+x_(6)=x_(2)+x_(7)=x_(3)+x_(8)=x_(4)+x_(9)=x_(5)+x_(10)=9` So, smallest number `1023489765` Largest number `9876501234` Total number `9xx8xx6xx4xx2xx1xx1xx1xx1xx1=3456` |
|