BDMO National Junior 2018/7

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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nahin munkar
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BDMO National Junior 2018/7

Unread post by nahin munkar » Tue Jan 08, 2019 1:53 pm

All possible $4$ digit numbers are created using $5,6,7,8$ and then sorted from smallest to largest. In the same manner, all possible $4$ digit numbers are created using $3,4,5,6$ and then sorted from smallest to largest. Then first number of the second type is subtract from first number of the first type, second number of the second type is subtract from second number of the first type and so on. What will be the summation of these difference (subtraction results) ?
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

BdMO National Junior problem 7

Unread post by samiul_samin » Thu Jan 10, 2019 1:49 am

You can get the problem here Short solution
there are total $4!=24$ permutaion for both numbers.
In every steps difference is $2222$
So answer is $2222×24=53328$

prottoy das
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Location:Sylhet

Re: BDMO National Junior 2018/7

Unread post by prottoy das » Wed Jan 16, 2019 11:22 pm

Probably it will be $4^4$ permutations. Because here, it is not said that a digit will be written one time in a number

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: BDMO National Junior 2018/7

Unread post by samiul_samin » Thu Jan 17, 2019 1:48 pm

prottoy das wrote:
Wed Jan 16, 2019 11:22 pm
Probably it will be $4^4$ permutations. Because here, it is not said that a digit will be written one time in a number
I really missed that :cry: :cry: :oops:

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