# Sum of elements of $A \cap(B \cup C)=400 \times 274$. Find the closed form of $C$.

Question:

$$A=\{x: x=3$$ digit natural number $$\}$$
$$B=\{x: x=9 k+2 ; k \in N\}$$
$${C}=\{{x}: {x}=9 {k}+I ; {k} \in {Z}\},$$ $$0 \leq {I}<9$$
Sum of elements of $$A \cap(B \cup C)=400 \times 274 .$$ Find the value of $$I$$

To be honest, I don't know how to start without making an assumption.
$$A \cap(B \cup C)=(A \cap B)\cup (A \cap C)$$ Assuming that $$(A \cap B)$$ and $$(A \cap C)$$ doesn't have any element common, one can find the sum of elements of $$(A \cap B)$$ and $$(A \cap C)$$ which would be equal to $$400 \times 274$$.

Sum of elements of $$(A \cap B)$$: $$\sum_{k=11}^{110}(9k+2)=54450+200=54650$$

Sum of elements of $$(A \cap C)$$:
Elements of $$C$$ are $$\cdots\cdots90+I,99+I,108+I\cdots990+I,999+I\cdots\cdots$$ Elements of $$A$$ are $$100,101,102\cdots 999$$ So possible elements of $$(A \cap C)$$ [By keeping $$0 \leq {I}<9$$ in mind] must be:$$99+I,108+I\cdots990+I\tag1\label{eq1}$$ So $$\sum_{k=11}^{110}(9k+I)=54450+100I$$ Then after adding and solving $$I=5$$ And by luck $$9k+2$$ and $$9k+5$$ does not have any element common from $$k=11$$ to $$k=110$$.

How can this be solved without making that risky assumption? Any alternate.
Seeing range in $$\eqref{eq1}$$ and guessing the elements of $$(A \cap C)$$ worked well in this question but there is a single value of $$I$$ so that won't work every time.

• If $I\ge2$, isn't $B\subseteq C$? Mar 2, 2021 at 6:49
• @Chrystomath I interpreted (perhaps wrongly) that $I \neq 2 \implies B ~\text{and}~ C~$ are disjoint. Assuming that my interpretation is correct, all that the OP needs to do is to break the problem into two cases [1] $I=2$ [2] $I \neq 2$, and explore each case separately. Mar 2, 2021 at 6:51
• You're right; $I$ is not inside the set condition; so it should be $C_I$ really. Mar 2, 2021 at 6:52

It is not by luck that those two sets do not have any elements in common. Assume $$9k+5=9\ell +2$$, where $$k,\ell$$ are integers. Then doing some algebra, we see $$9(\ell-k)=3$$. Do you see the problem here? This should help you make the assumption a lot less risky.
In words, $$A \cap(B \cup C)$$ contains all three-digits numbers which leave remainder $$2$$ and $$I$$ on division by $$9$$. For example,
$$A \cap B = \{ 99+2, \ldots, 990+2 \}$$
Clearly $$A \cap B$$ has $$110-11+1=100$$ elements whose sum is $$100 \cdot \frac{101+992}{2}=200 \cdot \frac{1093}{4}=200 \times 273.25$$
It is clear that $$I \neq 2$$ and $$A \cap B$$ and $$A \cap C$$ are disjoint. We observe there would be $$100$$ terms in $$A\cap C$$ too. The numbers have been chosen conveniently. The total sum can be written as $$\sum (9k+2)+(9k+I)=\sum 2(9k+2) + (9k+I) - (9k+2)$$ $$=2\sum (9k+2) + \sum (I-2)$$ $$=400 \times 273.25 + 100(I-2)$$
Therefore, $$100 (I-2) = 400 \, (274-273.25) \Rightarrow \boxed{I=5}$$