"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.
"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.
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.
$\displaystyle \text{We have: }\;N \;=\;\underbrace{222\hdots2}_{m\text{ digits}} \underbrace{000\hdots0}_{n\text{ digits}}\;=\;2004k $
$\displaystyle \text{This can be written: }\;N \;=\;2\underbrace{(111\hdots1)}_{m\text{ digits}} \cdot 10^n \;=\;2^2\cdot501\cdot k $
$\displaystyle \text{Divide by 2: }\;10^n\cdot(111\hdots1) \;=\;2\cdot501\cdot k $
$\displaystyle \text{Divide by 2: }\;5\cdot 10^{n-1}(111\hdots1) \;=\;501\cdot k$
It seems that a number of the form $\displaystyle (111\hdots1)$ is a multiple of 501.
And all we have to do is find it . . . or prove that it exists.
Anyone? .Anyone?