Let's find the smallest natural number such that in its 47 times, the last four digits read 1969.
I. Solution. The sought $N$ number, if it exists, has at most 4 digits. Indeed, the digits of the thousands and higher place values can only influence such digits in their multiples, not the last 4; thus, in the smallest multiple showing the desired ending, there is no thousands digit.

![](https://cdn.mathpix.com/cropped/2024_05_02_c36b3076648a4be58af1g-1.jpg?height=320&width=358&top_left_y=306&top_left_x=878)

We denote the digits of $N$ - from right to left - by $a, b, c, d$ and determine them by performing the multiplication in reverse. The 9 in the units place of the product is the units digit of the partial product $7 \cdot a$, so $a=7$, because (among the products of single-digit numbers with 7) only the 7-fold ends in 9. Thus, in the tens place column of the multiplication scheme (in the lower row), the 4 tens from $7 \cdot a = 49$ are carried over, and (in the upper row) the units digit of the product $4 \cdot a$, which is 8, and (in the lower row) the units digit of the product $7 \cdot b$. The already existing $4+8=12$ tens are completed to 6 tens in the product; and since only the 2-fold of 7 ends in 4, $b=2$.

Similarly, we obtain the other digits - in the hundreds and thousands columns, taking into account the carry-over from the column behind them: $c=1$, and finally $d=8$, so the sought number is $N=8127$, and $47N=381969$.

Remark. The last 3 digits of $N$ can be read from the common multiplication tables (the products of the numbers 2-99 with the numbers 2-999) - only the multiples of 47 in the form $10k+7$ end in 9, only the multiples of (100k +27) end in 69, and among them, only the 127-fold ends in 969. Thus, we only need to determine $d$. - However, our reasoning is faster than this search.

II. Solution. According to the requirement,

$$
47 N=1969+10000 P
$$

where $P$ is a natural number, and we are looking for the smallest such $N$ (and thus the smallest $P$). By division,

$$
\begin{aligned}
N & =41+212 P+\frac{42+36 P}{47}=41+212 P+6 \cdot \frac{7+6 P}{47}= \\
& =41+212 P+6 Q, \quad 47 Q=7+6 P,
\end{aligned}
$$

where $Q$ is also a natural number, because $P$ is positive, and 6 and 47 are relatively prime; we are looking for the smallest such $Q$, because then $P$ will also be the smallest among the suitable numbers. Similarly,

$$
P=8 Q-1-\frac{Q+1}{6}
$$

and the last term is an integer.

We immediately see that the smallest suitable value here is $Q=5$, with which $P=38$, and $N=8127$.[^0]


[^0]:    ${ }^{1}$ For example, O. Schade: Multiplication Table, 2nd edition, Műszaki Könyvkiadó, Budapest, 1964. - György Demény: Light Multiplier, Könnyűipari Kiadó, Budapest, 1952.