## Dhaka Higher Secondary 2011/8

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
Forum rules
Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### Dhaka Higher Secondary 2011/8

$N$ 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 $k + l$.

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm
Location: Khulna,Bangladesh.

### Re: Dhaka Higher Secondary 2011/8

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]
বড় ভালবাসি তোমায়,মা

Hasib
Posts: 238
Joined: Fri Dec 10, 2010 11:29 am
Location: খুলনা, বাংলাদেশ
Contact:

### Re: Dhaka Higher Secondary 2011/8

plz clear why a+b+c has $3^5$ choice?
A man is not finished when he's defeated, he's finished when he quits.

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm
Location: Khulna,Bangladesh.

### Re: Dhaka Higher Secondary 2011/8

i got the ans wrong
didn't notice the condition that all the numbers were distict.
বড় ভালবাসি তোমায়,মা

Mehfuj Zahir
Posts: 78
Joined: Thu Jan 20, 2011 10:46 am

### Re: Dhaka Higher Secondary 2011/8

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

Marzan
Posts: 2
Joined: Tue Feb 08, 2011 9:56 pm
Location: Mohammadpur, Dhaka

### Re: Dhaka Higher Secondary 2011/8

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
Posts: 2
Joined: Tue Feb 08, 2011 9:56 pm
Location: Mohammadpur, Dhaka

### Re: Dhaka Higher Secondary 2011/8

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???? )

bristy1588
Posts: 92
Joined: Sun Jun 19, 2011 10:31 am

### Re: Dhaka Higher Secondary 2011/8

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

I get this as the answer, What is the actual answer>??
Bristy Sikder

nafistiham
Posts: 829
Joined: Mon Oct 17, 2011 3:56 pm
Location: 24.758613,90.400161
Contact:

### Re: Dhaka Higher Secondary 2011/8

my answer is also $10$
i have gone case by case.
$\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0$
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm
Location: Dhaka, Bangladesh.
Contact:

### Re: Dhaka Higher Secondary 2011/8

তিহাম... তাহমিদের সল্যুশনে ভুল ছিল কিন্তু... খেয়াল করস... আরেকবার ভাব...
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes