investigate the divisibility of positive integers

Asked Aug 31 '20 at 12:14

Active Sep 05 '20 at 10:11

Viewed 77 times

I have an other math problem what I need your help:

Two people investigate the divisibility of positive integers. I call them person $1$ and $2$.

First person $1$ gives a digit $a$ and forms the number with the decimal representation $100a$.

Then person $2$ chooses a digit $b$. Now person $1$ shall form a number of the form $100ba, 100bba, 100bbba, ... . $

If person $1$ finds such a number, which has no common divisor greater than $1$ with $100a$, he has won, otherwise person $2$ wins.

Now I shall determine all digits $a$, by which choice person $1$ can secure his win.

Hint: $abcd$ is the positive integer, which has the digits $a, b, c$ and $d$ in the decimal representation from left to right.

I myself tried to approach the task mathematically, but did not get a real result. No matter which number I used for $a$ or $b$, the result had more than one common divisor.

edited Sep 05 '20 at 10:11

Alessio K

10,599

asked Aug 31 '20 at 12:14

Jonathan

Just to clarify, by "100a" you mean $1000+a$ ? So that if $a=8$ then "100a"= $1008$ ? – Matti P. Aug 31 '20 at 12:20
Yes that is right – Jonathan Aug 31 '20 at 12:29
And for example 100bba is 100773 if b is b =7 and a=3 – Jonathan Aug 31 '20 at 12:33
If, say, $a$ is even then all those numbers are even, yes? So all even digits fail for the first player. Similarly if $a$ is $5$, all the numbers are divisible by $5$. – lulu Aug 31 '20 at 12:33
That is right. So 1, 3 and 7 remain for a – Jonathan Aug 31 '20 at 12:37
And $9$, which is your best candidate (since $1009$ is a prime). – lulu Aug 31 '20 at 12:39
Yes and 9 that is right – Jonathan Aug 31 '20 at 12:40
But we forget, that we must find a number, which must not have a divisor with 100a except 1. – Jonathan Aug 31 '20 at 12:44
How can mathematically prove that it applies only to these numbers. – Jonathan Aug 31 '20 at 12:47
What have you tried? I gave you a simple road map to rule out most of the cases. You should easily be able to rule out $1,3,7$. That only takes a few seconds. That only leaves you with $9$. – lulu Aug 31 '20 at 12:50
Sorry but why only with 9. I am so bad in maths – Jonathan Aug 31 '20 at 12:52
Because it is a prime – Jonathan Aug 31 '20 at 12:53
Try each case $1,3,7$. a simple search shows that player $1$ loses in each case. – lulu Aug 31 '20 at 12:55
@lulu Do you mean seconds by a computation in the head or with the computer ? – Peter Aug 31 '20 at 13:02
@Peter With a machine. Should add: I don't think $9$ works either, though the counterexample is harder to find. $b=0$ yields a counterexample eventually (after all $10$ has finite order $\pmod {1009}$, $252$ as it happens, so $10^{255}+9\equiv 0 \pmod {1009}$. – lulu Aug 31 '20 at 13:04
In many cases (maybe all the reamining cases) , player $1$ will even be able to produce a prime. – Peter Aug 31 '20 at 13:08
Ok but is there a method that can prove that 9 ist the best result except the fact that it is a prime. And in the tsk it is talked about digits not only one. I am not sure. – Jonathan Aug 31 '20 at 13:54
I noticed, that for a number to be divisible by 3 or 9, its cross sum has to be divisible by 3 or 9. – Jonathan Aug 31 '20 at 14:04

investigate the divisibility of positive integers

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