Determine the digits of the product

$333 \ldots 33 \cdot 666 \ldots 66$

knowing that both factors are 25-digit numbers.
I. Solution: Using the conventional method of written multiplication (taking either factor as the multiplier), each partial product is the 26-digit number written with 26 digits, whose first digit is 1, the last digit is 8, and all intermediate digits are 9. Indeed, from the 6-fold of the 3 in the units place, 18, the lowest place value digit of each partial product is 8, and it is "carried over" as 1 to the 6-fold of the 3 in the next higher place value; from $1+3 \cdot 6=19$, we write 9, again "carrying over" 1, and this repeats 23 more times because each subsequent remainder is the same as the first; finally, after multiplying the leading 3 (counting from the right, the 25th), we also write the carried 1 as the first (counting from the right, the 26th) digit of the partial product.

Writing the partial products in the usual stepped manner, we start with 26 columns of digits in the first one, and with the next 25 partial products, the number of columns increases to $26+24=50$. In each of the first 25 columns, there is a single 1, and in addition (moving from left to right) $0,1,2, \ldots, 23,24$ nines, and in the next, the first 25 columns from the right, there is a single 8 and $0,1,2, \ldots, 24$ nines.

In the conventional addition, the "sum" of the first column (from the right) is 8 (and 0 is carried over), from the second column $0+(9+8)=17$, we write 7, and 1 is carried over, from the third column $1+9+(9+8)=27$, again 7 is written, and 2 is carried over. In the next $25-3=22$ columns, the sum increases by $9+1=10$ from step to step because we have one more 9, and the remainder increases by 1, so the written digit is always (in 24 columns from the tens place) 7. The remainder from the 24th column is 23, from the 25th is 24. Thus, the 26th column differs from the 25th in that it carries over 1 more remainder, and in that it contains a 1 instead of an 8, while the number of 9s is the same. Thus, the sum of the 26th column shows a change of $1-(8-1)=-6$ units compared to the 25th, the written digit is $7-6=1$, and the remainder remains 24. So we wrote 241, or carried over. In the 27th column, due to one less 9, the sum is $241-9=232$, from which we write 2, and 23 is carried over, which is 1 less than the previous one. Each subsequent column loses $9+1=10$ in the summands and the remainder, so the written digit is always 2, and no remainder is carried over from the last, the 50th column. Thus, the sought product $N$, as the sum of the partial products, is written with 50 digits: 24 2s, then a 1, then 24 7s, and finally an 8:

$$
N=\overbrace{222 \ldots 2}^{24 \text { digits }} 1 \overbrace{777 \ldots 7}^{24 \text { digits }} 8
$$

Gagyi Pálffy András (Bp. VIII., Széchenyi I. g. II. o. t.)

II. Solution: Divide the second factor by 3 and multiply the first factor by 3, so we can also obtain the desired product as the product of the numbers written with 25 9s and 25 2s, respectively. Writing the first factor in the form $10^{25}-1$,

$$
N=\overbrace{222 \ldots 2}^{25 \text { digits }} \cdot 10^{25}-\overbrace{222 \ldots 2}^{25 \text { digits }}
$$

and from this, by the conventional method of written subtraction, an 8 is placed in the first place from the right, and 7s in the next 24 places due to the carry-over; a 1 in the 26th place, where there is no digit in the subtrahend but there is still a remainder, and finally, in the next 24 places, the difference is the same as the minuend.

Simai László (Kisújszállás, Móricz Zs. g. II. o. t.)

Remarks. 1. It is natural to generally examine the product $N_{k}$ of the number with $k$ threes and the number with $k$ sixes. This can be written in the form of the transformations used in the second solution:

$$
N_{k}=3 \cdot \frac{10^{k}-1}{9} \cdot 6 \cdot \frac{10^{k}-1}{9}=\frac{2}{9}\left(10^{k}-1\right)^{2}
$$

We conjecture that the product is written with $2k$ digits, the first $k-1$ digits are 2s, the $k$-th digit is 1, the next $k-1$ digits are 7s, and the last digit is 8, i.e., that it is equal to the following number:

$$
N_{k}^{*}=2 \cdot \frac{10^{k-1}-1}{9} \cdot 10^{k+1}+1 \cdot 10^{k}+7 \cdot \frac{10^{k-1}-1}{9} \cdot 10+8
$$

A simple transformation shows that indeed $N_{k}^{*}=N_{k}$.

2. This conjecture can also be proven by the method of complete induction.

Bollobás Béla (Bp. V. Apáczai Csere J. gyak. g. II. o. t.)

3. Several papers almost mechanically calculated the product using the conventional method, not taking advantage of the fact that the digits of the factors are the same. The goal with this problem was for our solvers to think and reach the goal with less work; the text also hints at this: "determine the digits".