If we divide a number by 4, the remainder is 3, and if we divide it by 9, we get a remainder of 5. What is the remainder when we divide it by 36?
I. solution: Denoting the number in question by $N$,

$$
N=4 x+3=9 y+5
$$

where $x$ and $y$ are integers.

Reducing to zero and dividing by 4,

$$
x-2 y-\frac{y}{4}-\frac{1}{2}=0
$$

Denoting the integer $x-2 y$ by $(z+1)$ and multiplying by 4,

$$
4(z+1)-y-2=0
$$

or

$$
4 z-y+2=0
$$

from which

$$
y=4 z+2
$$

and thus

$$
N=9 y+5=9(4 z+2)+5=36 z+23
$$

It is evident from this that dividing $N$ by 36, the remainder is 23.

Pál Csonka (Bp. XI., József Attila g. IV. o. t.)

II. solution: Among the numbers between 0 and 36, the following four numbers give a remainder of 5 when divided by 9: 5, 14, 23, 32. However, only 23 satisfies our second requirement, which is to give a remainder of 3 when divided by 4.

Any number can be written in the form

$$
N=36 k+r
$$

where $k=0,1,2, \ldots$ and $r=0,1,2, \ldots, 35$.

Since 36 is divisible by both 9 and 4, $N$ when divided by 9 or 4 gives the same remainder as $r$. Therefore, $N-$ according to the above - can only meet our conditions if $r=23$, that is,

$$
N=36 k+23
$$

which clearly shows that the sought remainder is 23.

Sándor Kántor (Debrecen. Ref. g. IV. n. t.)