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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.

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

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

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