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John is thinking of a number $n$. He's willing to tell us that the number is close to $10000$ and in binary system it ends on $101$. In $7$ and $11$ system it ends on digit $2$ and the last two digits in the ternary numeral system are $21$. What is this number he is thinking of? I understand we need to create a set of congruences and solve it.

I have tried the following: Two of them are $x = 2, (mod 7)$ and $x = 2 (mod 11)$. That should be correct. The first one is $x = 5 (mod 8)$ and the last is $x = 7 (mod 9)$. I can see for this first one that $101$ is $5$ when converted to decimal system, but where do we get modulo $8$, from $2^3$? But why? Thank you for any clarification.

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  • $\begingroup$ For the same reason that a decimal number that ends in 65 is congruent to 65 modulus 100 as well as being congruenent to 5 modulus 10. $\endgroup$ Jan 18, 2019 at 12:42

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If your binary and ternary numerals are $\ b_nb_{n-1}\dots b_2b_1b_0\ $ and $\ t_mt_{m-1}\dots t_1t_0\ $, respectively, with $b_2=1, b_1=0, b_0=1, t_1=2\ $ and $\ t_0=1\ $, then, by definition, the numbers they represent are: \begin{eqnarray} B &=& b_n\,2^n + b_{n-1}\,2^{n-1} + \dots + b_3\,2^3 + 1\times2^2 + 0\times2 + 1\\ &=& 2^3\,\left(\,b_n\,2^{n-3}+b_{n-1}\,2^{n-4} + \dots + b_3\, \right) + 5\ \ \ \mbox{and}\\ T &=& t_m\,3^m + t_{m-1}\,3^{m-1} + \dots + t_2\,3^2 + 2\times3 + 1\\ &=& 3^2\,\left(\,t_m\,3^{m-2} + t_{m-1}\,3^{m-3} + \dots + t_2\,\right) + 7\ , \end{eqnarray}

from which you should be able to see that if you subtract $\ 5\ $ from $\ B\ $ the result is divisible by $\ 8\ $, and if you subtract $\ 7\ $ from $\ T\ $ the result is divisible by 9.

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