## An interesting problem

### 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.

\[\clubsuit\]

\[\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.**

### 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 $0$.

### Re: An interesting problem

are the last two $\equiv 0$ also?

"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please

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.### Re: An interesting problem

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

\[\clubsuit\]

\[\clubsuit\]

**A man is not finished when he's defeated, he's finished when he quits.**

### Re: An interesting problem

The answer is $3816547290$

ফেইসবুকের মত এইখানে যদি প্রব্লেমে লাইক দেওয়া যেত, তাহলে এই প্রব্লেমে লাইক দিতাম!!!

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

### 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.

(

(

**Important**: Please make sure that you have read about the**Rules, Posting Permissions and Forum Language**)### 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

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

### 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.

\[clubsuit\]

\[\overline{abcdefghi} \equiv 0(mod 9)\]

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.**