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I know that n = 7, so I should expect a remainder of 2 when divided by 5. However, finding the solution algebraically gives me a wrong remainder of 6. Please help me find my errors, or even find a better way to approach this problem. My work is shown below. $$$$ I know that a general solution involves finding n in terms of the equations: $$ n = 2k + 1$$ $$n = 3m +1$$ I try to make both equations have a divisor of 5 like this: $$\frac{5}{2}n = 5k + \frac{5}{2}$$ and $$\frac{5}{3}n = 5m + \frac{5}{3}$$ I try to find the value of n by noticing $n = \frac{5}{2}n -\frac{9}{10} \cdot(\frac{5}{3}n)$. $$n = (5k + \frac{5}{2}) - \frac{9}{10}\cdot(5m + \frac{5}{3})$$ This becomes $$n = (5k - 5m\cdot \frac{9}{10})+ (\frac{5}{2} -\frac{5}{3}\cdot\frac{9} {10})$$ When I simplify this I get $$n = 5(k-\frac{9m}{10})+1$$ I note that $1 = -5 + 6$, therefore $$n = 5(k-\frac{9m}{10}) -5 +6$$ So in the end $$n = 5(k-\frac{9m}{10} -1) +6$$ I get a remainder of 6 but I was expecting a remainder of 2. Thank you for your time.

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  • $\begingroup$ Are you implicitly assuming that $\frac{9m}{10}$ is an integer? Also, the two conditions you have - $n = 1 mod 2, n = 1 mod 3$ do not give a unique solution, i.e. 7 is not the only solution arising from these conditions $\endgroup$
    – P. J.
    Feb 3, 2021 at 5:52
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    $\begingroup$ Since $2$ and $3$ are both coprime to $5$, the initial modular identities don’t tell you anything about $x \bmod 5$. They do tell you by the Chinese remainder theorem that $x\equiv 1\bmod 6$ . $\endgroup$
    – Joffan
    Feb 3, 2021 at 5:55
  • $\begingroup$ I'm assuming that everything within the brackets is an integer. So I assume n = 5x + 6. Is this wrong? $\endgroup$ Feb 3, 2021 at 6:00
  • $\begingroup$ Yes, because when you assumed that $n =3m+1$, $m$ was any integer, not necessarily divisible by 10. Therefore, you cannot say with certainty that $\frac{9m}{10}$ is an integer. $\endgroup$
    – P. J.
    Feb 3, 2021 at 6:27
  • $\begingroup$ Thank you for this clarification $\endgroup$ Feb 3, 2021 at 6:34

3 Answers 3

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$n$ could be $1,7,13,19$ or $25$, so $n\pmod5$ could be $1,2,3,4$ or $0$.

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  • $\begingroup$ i.e. $\bmod 5\!:\ n = 1+6k \equiv 1+k\,$ takes all values, i.e. is surjective (onto). $\endgroup$ Feb 3, 2021 at 8:48
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So I'm a bit confused, $6k+1$ is a solution of $n mod2 =1$ and $n mod3 =1$. So the solution isn't unique unless I'm missing something. Once you tell me that it is 7 then the earlier hypotheses are unnecessary.

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  • $\begingroup$ Thank you for your answer. You've helped me realize that the solution cannot be unique. $\endgroup$ Feb 3, 2021 at 6:01
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Every odd integer is of the form $2k+1$.

When you say that $n = 3k+1$, you mean that $n$ is an odd integer which gives remainder $1$ when divided by $3$.

Thus, you mean that $n$ is of the form $6k+1$.

This means that:

$$n = 5k + k + 1$$ $$n = k+1(mod 5)$$

Thus the expected set of remainders must be:$$remainder = {0,1,2,3,4}$$

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  • $\begingroup$ Thank you so much! $\endgroup$ Feb 3, 2021 at 6:17
  • $\begingroup$ I am glad you understood $\endgroup$ Feb 3, 2021 at 6:19

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