## An interesting problem

For students of class 11-12 (age 16+)
Hasib
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### An interesting problem

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.

Masum
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Joined: Tue Dec 07, 2010 1:12 pm

### Re: An interesting problem

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 $0$.

Moon
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### Re: An interesting problem

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

learn how to write equations, and don't forget to read Forum Guide and Rules.

Hasib
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### Re: An interesting problem

@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.

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

### Re: An interesting problem

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

Zzzz
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### Re: An interesting problem

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.

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

### Re: An interesting problem

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
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### Re: An interesting problem

@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.