# Error in this Chinese Remainder Theorem problem with three congruence equations?

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.

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

$$3\cdot14\cdot4+1\cdot35\cdot1+2\cdot10\cdot5=168+35+100=\color{red}{303}=23\pmod{70}$$

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

$$3\cdot14\cdot4+1\cdot35\cdot1+2\cdot10\cdot5=168+35+100=303=23\pmod{70}$$

• 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. May 9 '14 at 14:52