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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, then

A

the digit in tens place for smallest number is 6

B

the digit in tens place for largest number is 3

C

total numbers of such numbers is 3456

D

total numbers of such numbers is 365

Text Solution

Verified by Experts

The correct Answer is:
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`
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