# Given n mod 2 = 1 and n mod 3 =1. Find n mod 5. How to solve algebraically knowing n = 7

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.

• 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 Feb 3, 2021 at 5:52
• 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 $x\equiv 1\bmod 6$ . Feb 3, 2021 at 5:55
• I'm assuming that everything within the brackets is an integer. So I assume n = 5x + 6. Is this wrong? Feb 3, 2021 at 6:00
• 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. Feb 3, 2021 at 6:27
• Thank you for this clarification Feb 3, 2021 at 6:34

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

• i.e. $\bmod 5\!:\ n = 1+6k \equiv 1+k\,$ takes all values, i.e. is surjective (onto). Feb 3, 2021 at 8:48

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.

• Thank you for your answer. You've helped me realize that the solution cannot be unique. Feb 3, 2021 at 6:01

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:$$remainder = {0,1,2,3,4}$$

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