# Find this seven digit phone number under certain conditions

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?

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 $$x=435\;\;\;\;\wedge \;\;\;\;y=8717$$

• How do you get $998\mid 39x+1$ from $998\mid 19999x+1$? – jochen May 18 at 16:39
• $19999x +1$ reduce by $20\cdot 998x$ – Aqua May 18 at 16:41