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Problem 618

AIME late

Combinatorics Difficulty 5.0 Find the answer

Example 9 Let the plane region $D$ be represented by $N(D)$ , which denotes the number of all integer points (i.e., points on the xoy plane where both coordinates $x$ and $y$ are integers) belonging to $D$ . If $A$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=10$ , $y=1$ (including the boundaries), and $B$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=1$ , $y=100$ (including the boundaries), then $N(A \cup B)+N(A \cap B)=$ $\qquad$
$(1992$ , Shanghai Senior High School Mathematics Competition)

Your answer

Official solution

Solution: Draw the figure in the Cartesian coordinate system, and it is easy to calculate from the figure
$\begin{array}{l} N(A)=1^{2}+2^{2}+\cdots+10^{2} \\ =\frac{1}{6} \times 10(10+1)(2 \times 10+1)=385, \\ N(B)=\left(101-1^{2}\right)+\left(101-2^{2}\right)+\cdots \\ +\left(101-10^{2}\right) \\ =101 \times 10-\left(1^{2}+2^{2}+\cdots+10^{2}\right) \\ =1010-385=625. \\ \end{array}$
$N(A \cap B)=10.$

By $N(A \cup B)=N(A)+N(B)-N(A \cap B)$ , we get
$\begin{array}{l} N(A \cup B)=385+625-10=1000. \\ \therefore N(A \cup B)+N(A \cap B)=1010. \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.