For numerous technical applications (e.g., electronic computing), numbers need to be expressed in the binary system - that is, as the sum of different powers of 2. Express the number 413 in binary.
I. solution. Taking the powers of 2 in sequence, $2^{8}=256$ is the last one that is still smaller than 413. Thus, this is the largest power that can still appear in the sum, since every power of 2 (and any positive number) is positive. This reduces the problem to expressing $413-2^{8}=157$ as the sum of different powers of 2. Similarly,

$$
\begin{array}{rlr}
2^{7}=128<157, & \text { and } & 157-128=29 \\
2^{4}=16<29, & \text { and } & 29-16=13, \\
2^{3}=8<13 & \text { and } & 13-8=5
\end{array}
$$

finally $5=2^{2}+1=2^{2}+2^{0}$. Thus,

$$
413=2^{8}+2^{7}+2^{4}+2^{3}+2^{2}+1
$$

writing out the intermediate, non-occurring powers as well,

$$
413=1 \cdot 2^{8}+1 \cdot 2^{7}+0 \cdot 2^{6}+0 \cdot 2^{5}+1 \cdot 2^{4}+1 \cdot 2^{3}+1 \cdot 2^{2}+0 \cdot 2+1
$$

finally, writing out the powers of the base number (2) and the additions in the manner of the place value notation of the decimal system,

$$
413=(110011101)_{2}
$$

Kálmán György (Debrecen, Péterfia úti ált. isk. VIII. o. t.)

II. solution. The expression we are looking for will clearly be of this form:

$$
413=a \cdot 2^{n}+b \cdot 2^{n-1}+c \cdot 2^{n-2}+\ldots+m \cdot 2^{2}+n \cdot 2+p
$$

where the values of the coefficients $a, b, c, \ldots, m, n, p$ can each be 0 or 1, and are 1 if the corresponding power of 2 appears in the sum, and 0 otherwise.

The last term on the right side can be factored out by 2, so

$$
413=2\left(a \cdot 2^{n-1}+b \cdot 2^{n-2}+c \cdot 2^{n-3}+\ldots+m \cdot 2+n\right)+p
$$

This means that the remainder of the division 413 : 2 gives the last digit of the expression we are looking for, $p$, and the quotient gives the expression in parentheses: $413=2 \cdot 206+1$, so $p=1$ and

$$
a \cdot 2^{n-1}+b \cdot 2^{n-2}+\ldots+m \cdot 2+n=206
$$

Now we only need to express the smaller number 206 in the binary system and write the digit $p=1$ at the end of the expression. Similarly, dividing 206 by 2, the last digit before the last digit of 413 is $n=0$, and

$$
a \cdot 2^{n-2}+\cdot 2^{n-3}+\ldots+m=103
$$

Continuing the procedure,
$103: 2=51$,
1
$6: 2=3$,
0
$51: 2=25$,
$25: 2=12$,
$12: 2=6$,
1
1
0
$3: 2=1$,
$1: 2=0$,
1
1

we obtained the digits of the expression we are looking for in reverse order from the remainders of the successive divisions: 413 = $(110011101)_{2}$.

Buócz Enikő (Miskolc, Kilián Gy. g. II. o. t.)