Number theory problem involving linear congruences

Question

What three positive integers, upon being multiplied by 3, 5, and 7 respectively and the products divided by 20, have remainders in arithmetic progression with common difference 1 and quotients equal to remainders?

My thought process is to solve the system of linear congruences

(1) $r \equiv3n_1 \pmod{20}$

(2) $r + 1 \equiv 5n_2 \pmod{20}$

(3) $ r + 2 \equiv 7n_3\pmod{20}$

If I could solve this for $r$, using the chinese remainder theorem, my thought process is that I could then find $n_1, n_2, n_3$ which work.

Plugging (1) into (2) gives $3n_1\equiv 5n_2-1 \pmod{20}$. Since the inverse of $3$ mod $20$ is $7$, I could isolate $n_1\equiv 7(5n_2-1) \pmod{20}$.

Now I have that $r\equiv3(7(5n_2-1))=105n_2-21$ which I could then plug into the third equation.

I now have $105n_2-21\equiv7n_3-2 \pmod{20} \Rightarrow 105n_2\equiv 7n_3+19 \pmod{20} $. However, since 105 does not have an inverse mod 20 since $105$ and $20$ are not coprime, I am unable to proceed from here.

Am I on the right track? What can I do from here to solve this problem?

I also feel I am disregarding the part of the question which says "and quotients equal to remainders".

Answer 1

If I understand the question correctly, one needs to solve the following: $$3n_1=20r+r,$$ $$5n_2=20(r+1)+(r+1),$$ and $$7n_3=20(r+2)+(r+2).$$ The first and third are obviously met. To solve for the second, just need to require that $r=5t-1,t\geq 1$. It follows that $$n_1=7(5t-1),n_2=21t,n_3=3(5t+1),t\geq 1.$$ If one further requires that $r+2<20$, then one can proceed to identify all solutions.

Answer 2

To answer as the OP says "involving linear congruences," suppose the three integers are $a$, $b$, and $c$. Then, by the Division Algorithm, using $r$ as the remainder:

\begin{align} 3a &=20r+r & 5b &=20(r+1)+(r+1) & 7c &=20(r+2)+(r+2) \\ 3a &=21r & &=21r+21 & &=21r+42 \end{align}

Then we can rewrite these as congruences with $r$:

\begin{align*} r &\equiv 0 \pmod 3 & r &\equiv -1 \pmod 5 & r &\equiv -2\pmod 7 \\ & & r &\equiv 4\pmod 5 & r &\equiv 5\pmod 7 \\ \end{align*}

We can solve for this system of three congruences using the Chinese Remainder Theorem (in this case, the \href{https://math.stackexchange.com/a/2147772/1335296}{Easy CRT}). First iteration:

\begin{align*} r &\equiv 0\pmod{3} & \iff r &\equiv 0 + 3 \bigg[\frac{{4}}{3}\pmod 5\bigg] \pmod {15}\\ r &\equiv 4 \pmod{5} & r &\equiv 3\cdot \bigg[\frac{{4+5}}{3}\pmod 5\bigg] \pmod{15}\\ & & r &\equiv 9 \pmod{15} \end{align*}

Second iteration:

\begin{align*} r &\equiv 9\pmod{15} & \iff r &\equiv 9 + 15 \bigg[\frac{{-4}}{15}\pmod 7\bigg] \pmod {105}\\ r &\equiv 5 \pmod{7} & r &\equiv 9+15\bigg[\frac{{45}}{15}\pmod 7\bigg] \pmod{105}\\ & & r &\equiv 9 + 15\cdot 3\pmod{105}\\ & & r &\equiv 9+45\pmod {105}\equiv 54 \pmod{105}\\ \end{align*}

So $r\equiv 54\pmod {105}$ Substitute back into the first set of equations above to find the three positive integers:

\begin{align*} 3a &=21r & 5b &=21r+21 & 7c &=21r+42 \\ &=21(54) & &=21(54)+21 & &=21(54)+42\\ &=1134 & &=1155 & &=1176 \end{align*}

Then verify that the numbers meet the criteria of the problem:

\begin{align*} 3a &=1134 & 5b &=1155 & 7c &=1176 \\ &=20(54)+54 & &=20(55)+55 & &=20(56)+56 \end{align*}