# Thread: Today's problem that I don't understand.

1. ## Today's problem that I don't understand.

"Show that there is a multiple of 2004 which has the form 22...200...0, i.e., a sequence of digits 2 followed by a sequence of digits 0."

These seem so easy after I see the answer, but I never know exactly what is being asked.

2. Hello, billym!

A fascinating problem . . . I haven't solved it yet.
But I have a start . . . maybe someone can follow through?

Show that there is a multiple of 2004 which has the form 22...200...0,
that is, a sequence of 2's, followed by a sequence of 0's.

$\text{We have: }\;N \;=\;\underbrace{222\hdots2}_{m\text{ digits}} \underbrace{000\hdots0}_{n\text{ digits}}\;=\;2004k$

$\text{This can be written: }\;N \;=\;2\underbrace{(111\hdots1)}_{m\text{ digits}} \cdot 10^n \;=\;2^2\cdot501\cdot k$

$\text{Divide by 2: }\;10^n\cdot(111\hdots1) \;=\;2\cdot501\cdot k$

$\text{Divide by 2: }\;5\cdot 10^{n-1}(111\hdots1) \;=\;501\cdot k$

It seems that a number of the form $(111\hdots1)$ is a multiple of 501.

And all we have to do is find it . . . or prove that it exists.

Anyone? .Anyone?

3. Apparently helpful hint:

Consider the 2005 “pigeons” 2, 22, 222, . . . , 22...2 and put them into holes according to their remainder upon division by 2004. Then argue as in here.

4. Originally Posted by billym
Apparently helpful hint:
Consider the 2005 “pigeons” 2, 22, 222, . . . , 22...2 and put them into holes according to their remainder upon division by 2004. Then argue as in here.
What a completely surprising application of the pigeon hole principle.

5. This was really nice !! Thanks