7.4 Vanya wrote the numbers $1,2,3, \ldots, 13$ in his notebook. He multiplied five of them by 3, and the rest by 7, then added all the products. Could the result have been 433?
Solution. Let's try to figure out the situation. Let A be the sum of the five numbers that need to be multiplied by 3, and B be the sum of the other eight numbers (which need to be multiplied by 7). Consider the sum after multiplication: 3A + 7B. We can write it as (taking into account that the sum of all thirteen numbers is 91):

\[
3(A+B)+4 B=3 \cdot 91+4 B=273+4 B
\]

We would like this sum to be equal to 433:

\[
273+4 B=433, \quad 4 B=160, \quad B=40
\]

Let's select eight numbers with a sum of 40. The sum \(1+2+\ldots+7+12=40\).

The sum of the other five numbers: \(8+9+10+11+13=51\).

We can check:

\[
3(8+9+10+11+13)+7(1+2+\ldots+7+12)=3 \cdot 51+7 \cdot 40=153+280=433
\]

Answer: it could.