An interesting problem

For students of class 11-12 (age 16+)
Hasib
Posts: 238
Joined: Fri Dec 10, 2010 11:29 am
Location: খুলনা, বাংলাদেশ
Contact:

An interesting problem

Unread post by Hasib » Sat Dec 11, 2010 6:32 pm

Solve this:
\[\begin{align*}a & \equiv 0(mod 1) \\
10a+b & \equiv 0(mod 2) \\
100a+10b+c & \equiv 0(mod 3) \\
1000a+100b+10c+d & \equiv 0(mod 4) \\
10000a+1000b+100c+10d+e & \equiv 0(mod 5) \end{align*}\]

it's continuas up to j. That means:
\[abcdefghij \equiv 0(mod10)\]

the all {a,b,c,d....j} are distinct.
\[\clubsuit\]
Last edited by Hasib on Sun Dec 12, 2010 3:31 pm, edited 3 times in total.
A man is not finished when he's defeated, he's finished when he quits.

User avatar
Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
Location: Dhaka,Bangladesh

Re: An interesting problem

Unread post by Masum » Sat Dec 11, 2010 8:17 pm

Just add two dollar signs before and after your mathematical signs and they will convert into LaTeX
One one thing is neutral in the universe, that is $0$.

User avatar
Moon
Site Admin
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Location: Dhaka, Bangladesh
Contact:

Re: An interesting problem

Unread post by Moon » Sat Dec 11, 2010 9:25 pm

are the last two $\equiv 0$ also?
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

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

Re: An interesting problem

Unread post by Hasib » Sat Dec 11, 2010 10:27 pm

@Moon vaia: Sorry, now the problem is correct fully!!

\[\clubsuit\]
A man is not finished when he's defeated, he's finished when he quits.

User avatar
Avik Roy
Posts: 156
Joined: Tue Dec 07, 2010 2:07 am

Re: An interesting problem

Unread post by Avik Roy » Sun Dec 12, 2010 1:07 pm

The answer is $3816547290$
ফেইসবুকের মত এইখানে যদি প্রব্লেমে লাইক দেওয়া যেত, তাহলে এই প্রব্লেমে লাইক দিতাম!!! :D
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

User avatar
Zzzz
Posts: 172
Joined: Tue Dec 07, 2010 6:28 am
Location: 22° 48' 0" N / 89° 33' 0" E

Re: An interesting problem

Unread post by Zzzz » Sun Dec 12, 2010 2:47 pm

অভীক ভাই, পুরাটা কি logic দিয়ে বাইর করছেন ? করলে please একটু ব্যাখ্যা করেন। আমি বের করছিলাম কিছুটা logic আর কিছুটা trial and error দিয়ে।
Every logical solution to a problem has its own beauty.
(Important: Please make sure that you have read about the Rules, Posting Permissions and Forum Language)

User avatar
Avik Roy
Posts: 156
Joined: Tue Dec 07, 2010 2:07 am

Re: An interesting problem

Unread post by Avik Roy » Sun Dec 12, 2010 3:30 pm

কিছুটা trial and error আমারও লেগেছে। আমি সমাধানের হিন্টস দিচ্ছি, দেখ কাজে লাগাতে পার কিনা।

1) $j=0,e=5$ are easy to deduce
2) Each trio of $\overline{abc}$, $\overline{def}$ and $\overline{ghi}$ will consist of one $0(mod 3)$, one $1(mod 3)$ and $2(mod 3)$ digit.
3) $b,d,f,h$ are even. Using divide by $4$ condition and earlier findings, deduce that $b=8, d=6, f=4, h=2$
4) divide by $8$ condition leaves us with $g=3 or, 7$
5) divide by 7 condition leaves us with the relation $5c+3a+3g \equiv 0(mod 7)$
6) Plugging in available combinations and rejecting those that fail to follow (5), we get the desired solution
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

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

Re: An interesting problem

Unread post by Hasib » Sun Dec 12, 2010 3:43 pm

@Avik da: same problem. Jst i change the problem a bit. The main problem was upto i. That means upto:
\[\overline{abcdefghi} \equiv 0(mod 9)\]
And none of {a,b,c....i} are 0.

The problem is from The Art and craft of problem solving.

\[clubsuit\]
A man is not finished when he's defeated, he's finished when he quits.

Post Reply