From GMB 11/2025, page 554. In translation :
"E:17346. A natural number $n$ gives when divided by 7 the remainder 4
and when divided by 9 the remainder 5. Find out what remainder is
obtained when dividing $n$ by 63.
[Author] Ștefan GOBLEJ, Curtea de Argeș"
ANSWER CiP
32
Solution CiP
Let $a$ and $b$ be the quotients of the divisions of $n$ by 7 and 9 respectively. Then
$n=7\cdot a+4,\;\;\;n=9\cdot b+5 \;\;\;\;\;a,\;b\in \mathbb{N}\tag{1}$
The equation $7\cdot a+4=9\cdot b+5$ is equivalent to
$7\cdot a-9\cdot b=1\;\;\;\;\;\;a,b \in \mathbb{N} \tag{2}$
Noting that $7\cdot 4-9\cdot 3=28-27=1$ , equation (2) has a particular solution $a_0=4,\;b_0=3$.
So the general solution of this [linear Diophantine] equation is
$a=4+9\cdot k,\;\;b=3+7\cdot k,\;\;\;k \in \mathbb{N} \tag{3}$
Substituting the number $a$ from (3) into (1) we obtain $n=7\cdot (4+9\cdot k)+4$ so
$$n=63 \cdot k+32$$
and from this it follows that the remainder of dividing the number $n$ by 63 is 32.
$\blacksquare$
Remark CiP We could have thought like that too, without the theory of Diophantine equations :
$7\cdot a\overset{(2)}{=}1+9\cdot b=1+2\cdot b+7\cdot b\;\Rightarrow\; 7\mid 1+2\cdot b\;\Rightarrow\; 2\cdot b+1=7\cdot k_1,\;k_1 \in \mathbb{N}$. But in the equality $2\cdot b=7\cdot k_1-1$ we must have $k_1=odd\;number,\;\; k_1=2\cdot k+1$ so $2\cdot b=7(2\cdot k+1)-1=14 \cdot k+6$ and from here we obtain $b=7\cdot k+3$ that is precisely (3).
COPILOT's solution :
We write the numbers that give remainder of 4 when divided by 7 :
4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74, ...
We check which of them gives a remainder of 5 when divided by 9 :
4 mod 9 = 4
11 mod 9 = 2
18 mod 9 = 0
25 mod 9 =7
32 mod 9 =5 : BINGO
(CiP note : After at most nine (=9) attempts, the result is definitely found, whether it exists or not.)
So the number that satisfies both conditions is $n \equiv 32\;\;(mod\;63)$ , hence the remainder when dividing $n$ by 63 is 32.
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