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Re: Dhaka Higher Secondary 2011/8

Posted: Wed Dec 14, 2011 1:29 pm
by bristy1588
Can Aanyone tell me why i m wrong and what is the correct answer?

Re: Dhaka Higher Secondary 2011/8

Posted: Wed Dec 14, 2011 1:39 pm
by nafistiham
হায় হায়! ফাহিম ভাইয়ারও এই উত্তর আসছে । আমি কয়েকজনকে তো এভাবেই দেখালাম । কিন্তু কিভাবে হবে ? এই টপিকে তো কোথাও সমাধানটা নাই ।

Re: Dhaka Higher Secondary 2011/8

Posted: Wed Dec 14, 2011 1:48 pm
by bristy1588
nafistiham wrote:হায় হায়! ফাহিম ভাইয়ারও এই উত্তর আসছে । আমি কয়েকজনকে তো এভাবেই দেখালাম । কিন্তু কিভাবে হবে ? এই টপিকে তো কোথাও সমাধানটা নাই ।
Tiham, Fahim er Solution ta ki??

Re: Dhaka Higher Secondary 2011/8

Posted: Thu Dec 15, 2011 1:08 am
by Labib
I've got $6(115\cdot 146+11\cdot 147)$.
Really confused... :?
Please post your full solutions.... I'll post mine tomorrow.

Re: Dhaka Higher Secondary 2011/8

Posted: Thu Dec 15, 2011 1:58 pm
by nafistiham
unfortunately, my solution is so long that i hardly can even write it in my exercise book.so, i am trying to shorten it.whenever i can, i'll post

Re: Dhaka Higher Secondary 2011/8

Posted: Thu Dec 15, 2011 3:03 pm
by Labib
Here's my proof... (short version)
There are $180$, '$3$' digit numbers divisible by $3$,having none of its digits same.(All the digits are non-zero)
Let me assign them to the set $S_1$.
Again there are $80$, '$3$' digit numbers divisible by $6$ and having none of its digits same.(All the digits are non-zero)
These $80$ are also the element of $S_1$.

Now, if the number $N= \overline{abcdefghi}$
Then $\overline{def}$ can be chosen in $80$ ways and there will be $179$ ways left to choose $\overline{abc}$.

For each of the $\overline{abcdef}$, $\overline{ghi}$ can be arranged in $6$ ways yielding the conclution that,
$N$ can have $6(80\cdot 179)$ different values.

Re: Dhaka Higher Secondary 2011/8

Posted: Thu Dec 15, 2011 3:11 pm
by Labib
Alright! done editing and the solution's ready to go...
Waiting for others to post their solution.

Re: Dhaka Higher Secondary 2011/8

Posted: Tue Jan 07, 2014 8:38 pm
by mission264
i've got $N=2^6 \cdot 3^2 \cdot 37$

Re: Dhaka Higher Secondary 2011/8

Posted: Fri Jan 10, 2014 12:42 am
by Labib
I think the solution should be $2^6.3^2.37$. But it doesn't seem to fit with the problem statement. :?
Ignore my previous stupid posts btw. :mrgreen: