I3.1 Of the positive integers from 1 to 1000 , including 1 and 1000 , there are $\alpha$ of them that are not divisible by 5 or 7 . Determine the value of $\alpha$.

I3.2 Determine the value of $\beta=1^{2}-2^{2}+3^{2}-4^{2}+\cdots+(-1)^{\alpha}(\alpha+1)^{2}$.
I3.3 If $\gamma$ is the remainder of $\beta$ divided by the $1993^{\text {rd }}$ term of the following sequence: $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5, \cdots$. Determine the value of $\gamma$.

I3.4 In the figure below, $B E=A C, B D=\frac{1}{2}$ and $D E+B C=1$.
If $\delta$ is $\gamma$ times the length of $E D$, determine the value of $\delta$.
Numbers divisible by $5: 5,10,15, \cdots, 1000$, there are 200 numbers
Numbers divisible by $7: 7,14,21, \cdots, 994$, there are 142 numbers
Numbers divisible by $35: 35,70, \cdots, 980$, there are 28 numbers
Numbers divisible by 5 or $7=200+142-28=314$
Numbers that are not divisible by 5 or $7=1000-314=686$
Remark: The original question is:
Of the positive integers from 1 to 1000 , including 1 and 1000 , there are $\alpha$ of them that are not divisible by 5 or not divisible by 7 . Determine the value of $\alpha$.
$$
\begin{array}{l} 
1^{2}-2^{2}+3^{2}-4^{2}+\cdots+685^{2}-686^{2}+687^{2} \\
=\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right)+\cdots+\left(685^{2}-686^{2}\right)+687^{2} \\
=(1+2)(1-2)+(3+4)(3-4)+\cdots+(685+686)(685-686)+687^{2} \\
=-3-7-\cdots-1371+687^{2}(\text { sum of } 343 \text { terms of A.P., common difference }=-4) \\
= \frac{-3-1371}{2} \times 343+471969 \\
= 236328 \\
1+2+3+\cdots+62=\frac{1+62}{2} \times 62=1953 \text { and } 1993-1953=40<63
\end{array}
$$

The $1993^{\text {rd }}$ term of the sequence is 63 .
$236328 \div 63$, by division, the remainder is 15 .
Remark: The original question is:
Determine the remainder of $\beta$ divided by the $1993^{\text {rd }}$ term of the following sequence: $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5, \cdots, \gamma$ is not mentioned.
Let $D E=x, B E=y$
Then $A C=y, B C=1-x$
It is easy to show that $\triangle B E D \sim \triangle B C A$ (equiangular)
$$
\begin{array}{l}
\left.\frac{D E}{B E}=\frac{A C}{B C} \text { (cor. sides, } \sim \Delta^{\prime} \mathrm{s}\right) \Rightarrow \frac{x}{y}=\frac{y}{1-x} \Rightarrow y^{2}=x(1-x) \\
B E^{2}+D E^{2}=B D^{2} \text { (Pythagoras' theorem) } \\
y^{2}+x^{2}=\frac{1}{4} \Rightarrow x(1-x)+x^{2}=\frac{1}{4} \Rightarrow x=\frac{1}{4} \\
\delta=\frac{15}{4}
\end{array}
$$