Dhaka-2 Secondary 2013 / 6

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Labib
Posts:411
Joined:Thu Dec 09, 2010 10:58 pm
Location:Dhaka, Bangladesh.
Dhaka-2 Secondary 2013 / 6

Unread post by Labib » Mon Jan 13, 2014 7:37 pm

The least consecutive ten numbers, all greater than $10$, are determined that are
respectively divisible by the numbers $1$ through $10$. Write down the smallest
number among these $10$.
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Labib
Posts:411
Joined:Thu Dec 09, 2010 10:58 pm
Location:Dhaka, Bangladesh.

Re: Dhaka-2 Secondary 2013 / 6

Unread post by Labib » Mon Jan 13, 2014 8:28 pm

Here's how I approached the problem.
The problem asks for an integer $(x+1)$ such that $x+i\equiv 0$ (mod $i$) for $i = 1, 2, ..., 10$.
We know, $i\equiv 0$ (mod $i$). So, $x+i\equiv i$ (mod $i$).
So, $x\equiv 0$ (mod $i$) for $i = 1, 2, ..., 10$.
That means, $x$ is a multiple of all the integers from $1$ through $10$.
The smallest value of $x$ is the $LCM$ of the integers from $1$ through $10$.
Can you solve the problem from here? ;)
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.


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

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