Origin - p5 - Example 5

I'm querying a possible error, thence I show the pdf as is. Is the 3 underlined in red supposed to be 2? scilicet, the last line should be $n = 2 \times 5 \times 7 $? Notation hails from here.

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  • $\begingroup$ Solution is $x\equiv 23 \pmod{70}$, so I'd say you're right. $\endgroup$ May 6 '14 at 11:54
  • $\begingroup$ @DanielFischer thanks. my answer still doesn't match. i showed more work. $\endgroup$ May 6 '14 at 14:02
  • $\begingroup$ $3\cdot 14\cdot 4 + 1\cdot 35\cdot 1 + 2\cdot 10 \cdot 5 = 168 + 35 + 100 = 303 = 4\cdot 70 + 23$, that last seems to be an arithmetical mistake. $\endgroup$ May 6 '14 at 14:06

The error is in your last line. The addition is wrong


Some ideas:

$$\begin{align*}2x=1\pmod5&\iff x=2^{-1}\pmod5&\iff& x=3\pmod 5\\{}\\ 3x=9\pmod 6&\iff x=3\pmod 2&\iff&x=1\pmod 2\\{}\\ 4x=1\pmod7&\iff x=4^{-1}\pmod7&\iff&x=2\pmod7\end{align*}$$

Observe that equations (1)-(2) above already show the number must end in $\;3\;$, and together with equation (3) we get the solution $\;23\;$ (modulo $\;5\cdot2\cdot7=70\;$, of course)

If you want to apply the CRT (in fact, one of its proofs), then as follows is one way:

$$\begin{align*}14^{-1}\pmod 5&=4\\{}\\ 35^{-1}\pmod2&=1\\{}\\ 10^{-1}\pmod7&=5\end{align*}$$

Thus, a solution modulo $\;7\cdot2\cdot5=70\;$ is


  • $\begingroup$ Can you please just write in your answer that the error was in my last line? I foozled the addition in the last line. thanks for another solution though. $\endgroup$ May 9 '14 at 14:52

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