Chittagong BdMO 2014: Juior 9

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atiab jobayer
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Chittagong BdMO 2014: Juior 9

Unread post by atiab jobayer » Fri Dec 05, 2014 10:46 am

How many six digit integers can be formed so that the number formed by the first, second and fourth digits (counting from left) as well as the other number formed by the third, fifth and sixth digits is divisible by 11? It is required that the third and fourth digits are different
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tanmoy
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Re: Chittagong BdMO 2014: Junior 9

Unread post by tanmoy » Fri Dec 05, 2014 12:01 pm

What is Juior :o I think that is Junior :)
"Questions we can't answer are far better than answers we can't question"

tanmoy
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Re: Chittagong BdMO 2014: Juior 9

Unread post by tanmoy » Fri Dec 05, 2014 12:55 pm

The answer is:$5913$.Suppose,the numbers are of the form $abcdef$.So,$abd$ and $cef$ are divisible by $11$.There are $81$ three digit numbers which are divisible by $11$.Suppose,the number $abd$ is $110$.Then,the number $cef$ can be $110 to 990=81$ numbers.But if $abd$ is $121$,then,there are $72$ choices for $cef$.$\because$,$c$ and $d$ can't be same,so,the three digit numbers which start with $1$ and divisible by $11$ can't be counted.So,there are $81+72\times 8=657$ six digit integers can be formed when the six digit integers start with $1$.$\therefore$ Total $657\times 9=5913$ integers can be formed. :D
"Questions we can't answer are far better than answers we can't question"

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