This was a problem that was recently asked at a competition I attended:
I have a seven-digit phone number that satisfies the following property: taking the last four digits and placing them in the front creates a number that is one more than twice my original number. What is my phone number?
The problem essentially asks us to find an integer $\overline{a_0a_1a_2a_3a_4a_5a_6}$ such that $$2\cdot\overline{a_0a_1a_2a_3a_4a_5a_6} + 1 = \overline{a_3a_4a_5a_6a_0a_1a_2},$$ where $0 \leq a_i \leq 9$ for each $i$. Our team couldn't solve this problem during the competition; unfortunately, even after the competition has ended, we still are unable to procure a solution to this problem. We've tried taking the number and reducing it in different modulos with little success; we also tried setting up a Diophantine equation with $a = \overline{a_0a_1a_2}$ and $b = \overline{a_3a_4a_5a_6}$, with little success as well. How would you do this problem?
Thank you in advance for your help.
Let $x=\overline{a_0a_1a_2}$ and $y=\overline{a_3a_4a_5a_6}$, then we get $$2(x10^4+y)+1 = y10^3+x$$
so we get $$ 19999x+1 = 998y\implies 998\mid 19999x+1 \;\;\wedge \;\;x\leq 498$$
so $$ 998\mid 39x+1$$
Since $100\leq x\leq 498$ we have
$$3901\leq 39x+1\leq 19423$$ so $$39x+1=998z,\;\;\;\;\;z\in \{4,5,...,19\}$$
So $39 \mid 16z+1$ and since $65\leq 16z+1\leq 305$
$$16z+1\in\{78,117,156,195,234,273\}$$
we get $$16z\in\{77,116,155,194,233,272\}$$
so $z=17$ and $$x=435\;\;\;\;\wedge \;\;\;\;y=8717$$