Dhaka Higher Secondary 2011/8

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Higher Secondary 2011/8

Unread post by BdMO » Fri Jan 28, 2011 10:30 pm

$N$ represents a nine digit number each of whose digits are different and nonzero. The number formed by its leftmost three digits is divisible by $3$ and the number formed by its leftmost six digits is divisible by $6$. It is found that $N$ can have $2^k3^l$ different values. Find the value of $k + l$.

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Tahmid Hasan
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Re: Dhaka Higher Secondary 2011/8

Unread post by Tahmid Hasan » Sat Jan 29, 2011 7:05 pm

i first pick a number $\overline {abcdefghi}$
here $6 \mid a+b+c+d+e+f$
also $3 \mid a+b+c$
so$ 3 \mid d+e+f$
thus$6 \mid d+e+f$
now i compute the choices for $a+b+c$
it's $3^5$.and then i compute the choices for $d+e+f$.
it's $3^32^2$.
by multiplying them i get the total choices.it's $3^82^2$.
so$k+l=8+2=10$[\hide]
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Re: Dhaka Higher Secondary 2011/8

Unread post by Hasib » Sun Jan 30, 2011 10:42 pm

plz clear why a+b+c has $3^5$ choice?
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Tahmid Hasan
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Re: Dhaka Higher Secondary 2011/8

Unread post by Tahmid Hasan » Sun Jan 30, 2011 10:51 pm

i got the ans wrong :(
didn't notice the condition that all the numbers were distict.
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Mehfuj Zahir
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Re: Dhaka Higher Secondary 2011/8

Unread post by Mehfuj Zahir » Sun Jan 30, 2011 11:05 pm

You have forgotten about last 3 differnts digit.They have also 6 permutation.Then you will get the correct ans k+l=12

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Re: Dhaka Higher Secondary 2011/8

Unread post by Marzan » Tue Feb 08, 2011 11:42 pm

Let me show how I advanced..... 1st I divided the numbers into 3 groups using mod i.e 0 mod 3, 1mod 3 and 2 mod 3. According to divisibility rule, numbers made up by 1st 3 digits/3 & numbers made up by 2nd 3 digits/3 & 2 both. so, 6th digit must be even. 2/4/6 or 8. for 2 mod 3 digits for the 2nd group,we can have 2x2! permutations and so we have 2x3! permutations for the leftmost 3 digits... for 0 mod 3 & 1 mod 3 digits for the 2nd group we can have 2! permutations each for the 2nd group and 2x3! permutations for the leftmost 3 digits...

Now let's consider that the 2nd group of digits be filled up with 0,1 & 2 mod 3 digits each. so for 2 mod 3 digits we have 2 x 3C1 x 3C1 x 2! permutations and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st three digits.

for 0 mod 3 and 1 mod 3 we have 3C1 x 3C1 x 2! permutations for the 2nd group and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st group respectively.....

multiplying the related permutations and adding them we have
4x12 + 36x48 + 2x12 + 18x48 + 2x12 + 18x48
= 2^5 x 3 x 37!!!

(That's the problem with my solution... Here comes a prime number!!! I'm kinda sure that my method's right. Can anyone help me in this case? Is my solution wrong or is the ques wrong???? )

Marzan
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Re: Dhaka Higher Secondary 2011/8

Unread post by Marzan » Tue Feb 08, 2011 11:44 pm

Let me show how I advanced..... 1st I divided the numbers into 3 groups using mod i.e 0 mod 3, 1mod 3 and 2 mod 3. According to divisibility rule, numbers made up by 1st 3 digits/3 & numbers made up by 2nd 3 digits/3 & 2 both. so, 6th digit must be even. 2/4/6 or 8. for 2 mod 3 digits for the 2nd group,we can have 2x2! permutations and so we have 2x3! permutations for the leftmost 3 digits... for 0 mod 3 & 1 mod 3 digits for the 2nd group we can have 2! permutations each for the 2nd group and 2x3! permutations for the leftmost 3 digits...

Now let's consider that the 2nd group of digits be filled up with 0,1 & 2 mod 3 digits each. so for 2 mod 3 digits we have 2 x 3C1 x 3C1 x 2! permutations and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st three digits.

for 0 mod 3 and 1 mod 3 we have 3C1 x 3C1 x 2! permutations for the 2nd group and 2C1 x 2C1 x 2C1 x 3! permutations for the 1st group respectively.....

multiplying the related permutations and adding them we have
4x12 + 36x48 + 2x12 + 18x48 + 2x12 + 18x48
= 2^5 x 3 x 37!!!

(That's the problem with my solution... Here comes a prime number!!! I'm kinda sure that my method's right. Can anyone help me in this case? Is my solution wrong or is the ques wrong???? :?: )

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bristy1588
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Re: Dhaka Higher Secondary 2011/8

Unread post by bristy1588 » Tue Nov 29, 2011 3:02 am

${24*(2^5*3^3+2^3*3)}$

I get this as the answer, What is the actual answer>??
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nafistiham
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Re: Dhaka Higher Secondary 2011/8

Unread post by nafistiham » Tue Dec 13, 2011 1:57 pm

my answer is also $10$
i have gone case by case.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Labib
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Re: Dhaka Higher Secondary 2011/8

Unread post by Labib » Wed Dec 14, 2011 12:13 am

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