7.5. There are p sticks of lengths $1,2,3, \ldots$, p (cm), from which an equilateral triangle needs to be formed. Can this be done if a) p $=100$; b) p = 99? (The sticks cannot be broken, all p sticks must be used.)
7.5. There are n sticks of lengths 1, 2, 3, ..., n (cm), from which an equilateral triangle needs to be formed. Can this be done if a) n = 100; b) n = 99? (The sticks cannot be broken, and all sticks must be used.)

Answer: a) no; b) yes. Hint: Clearly, the total length of all sticks must be divisible by 3 in order to distribute them into three equal-length parts. a) But the sum $1+2+\ldots+100$ is not divisible by 3 (to verify this, it is not necessary to calculate the sum $1+2+\ldots+100=101 \cdot 50$; the 100 numbers can be divided as follows: $(1+2+3)+(4+5+6)+\ldots+(97+98+99)+100$, and in all triples the sum is divisible by 3, but 100 is not divisible by 3). b) We will divide all 99 numbers into consecutive nines $(1+2+\ldots+9)+(10+11+\ldots+18)+\ldots+(91+92+\ldots+99)$. Each such nine of the form $9 k+1, 9 k+2, \ldots 9 k+9$ will be divided into three groups with the same sum, namely: $9 k+1+9 k+5+9 k+9=9 k+2+9 k+6+9 k+7=9 k+3+9 k+4+9 k+8$. By collecting the sticks of each of the three groups for different $k=0,1, \ldots, 10$, we will obtain three equal total lengths.