# Intermediate Modular Arithmetic Strategy And Questions [closed]

I have these problems and I know how to solve ones with two equations but not more. Can anyone provide solutions? Thanks in advance!

Find the smallest positive integer that satisfies the system of congruences \begin{align*} N &\equiv 1 \pmod{7}, \\ N &\equiv 7 \pmod{13}, \\ N &\equiv 13 \pmod{20}. \end{align*}

Find the smallest positive $N$ such that \begin{align*} N &\equiv 6 \pmod{12}, \\ N &\equiv 6 \pmod{18}, \\ N &\equiv 6 \pmod{24}, \\ N &\equiv 6 \pmod{30}, \\ N &\equiv 6 \pmod{60}. \end{align*}

How many positive integers less than or equal to $6\cdot7\cdot8\cdot9$ solve the system of congruences \begin{align*} m &\equiv 5 \pmod{6}, \\ m &\equiv 4 \pmod{7}, \\ m &\equiv 3 \pmod{8}, \\ m &\equiv 3 \pmod{9}. \end{align*}

## closed as off-topic by Matthew Conroy, iadvd, Math1000, user219577, user91500Sep 9 '16 at 6:02

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• Nobody will provide you with solutions, but will be glad to offer assistance if you tell us what you have tried. – Brandon Thomas Van Over Sep 8 '16 at 23:43
• Do you know about the Chinese remainder theorem? – Robert Soupe Sep 9 '16 at 0:43
• However, for the second problem, the answer's probably $\textrm{lcm}(12, 18, 24, 30, 60) + 6$. – Robert Soupe Sep 9 '16 at 0:44

For problem 1:

$$N=7p+1$$ $$7p+1 \equiv 7 \pmod{13}$$ $$7p \equiv 6 \pmod{13}$$ $$p \equiv 12 \pmod{13}$$

(This is calculated by finding the inverse of 7 mod 13).

$$N \equiv 7*12+1 \pmod{13*7}$$

$$N \equiv 85 \pmod{91}$$

$$N = 91q + 85$$

$$N \equiv 13 \pmod{20}$$

$$91q + 85 \equiv 13 \pmod{20}$$

$$91q \equiv 8 \pmod{20}$$

$$11q \equiv 8 \pmod{20}$$

$$q \equiv 8 \pmod{20}$$

$$N \equiv 91*8+85 \pmod{91*20}$$

$$N \equiv 813 mod \pmod{1820}$$

$$N = 813$$

For problem 2, $N = 6$ is the obvious solution.

For problem 3, we see that it is not possible to have $N \equiv 5 \pmod{6}$ and $N \equiv 3 \pmod{9}$ (first suggests $N$ is not divisible by 3, second suggests $N$ is divisible by 3). So, here there are 0 valid solutions.