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$(2)(4)_b$ represents a number written in base $b$. For example $21_3 = 7$. How to find the smallest $b$ such that $(17)(3)_b$ is a multiple of $7$?

Assuming,$(17)$ as the single digit,I computed the smallest base to be $32$.Unfortunately,there is no way I could confirm this answer,so I am asking it here.Am i right?

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    $\begingroup$ 32? If $b=32$, then $(17)(3)_b=17\times32+3=547$, which is not a multiple of 7. $\endgroup$ – Gerry Myerson Oct 20 '11 at 2:05
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We must have

$$17b + 3 \equiv 3(b + 1)\equiv 0\ \pmod 7\Longrightarrow b \equiv 6\ \pmod 7$$

So we can have $b = 6, 13, 20, ...$. The 17 suggests that b > 17, the smallest multiple of which is 20.

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    $\begingroup$ The $17$ doesn’t just suggest that $b>17$: it requires it. $\endgroup$ – Brian M. Scott Oct 20 '11 at 0:14
  • $\begingroup$ Could you explain $17b + 3 \equiv 3(b + 1)\equiv 0\ \pmod 7\Longrightarrow b \equiv 6\ \pmod 7\space$? I can deduce $17b + 3 \equiv 0 \pmod 7 $ but the rest I couldn't follow. $\endgroup$ – Quixotic Oct 20 '11 at 0:20
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    $\begingroup$ @TretwickMarian Note that $17\equiv 3 \pmod{7}$, so $17b+3\equiv 3b+3=3(b+1)\pmod{7}$. Since $\gcd(3,7)=1$, if $3(b+1)\equiv 0\pmod{7}$, then $b+1\equiv 0\pmod{7}$ (if $7$ divides $3(b+1)$, then it has to divide $b+1$). If $b+1\equiv 0\pmod{7}$, then $b\equiv -1\pmod{7}$, and $-1\equiv 6\pmod{7}$. So $b\equiv 6\pmod{7}$. $\endgroup$ – Arturo Magidin Oct 20 '11 at 0:28

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