$A=\{x: x=3$ 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.

  • $\begingroup$ If $I\ge2$, isn't $B\subseteq C$? $\endgroup$ Mar 2, 2021 at 6:49
  • $\begingroup$ @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. $\endgroup$ Mar 2, 2021 at 6:51
  • $\begingroup$ You're right; $I$ is not inside the set condition; so it should be $C_I$ really. $\endgroup$ Mar 2, 2021 at 6:52

2 Answers 2


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, $$100 (I-2) = 400 \, (274-273.25) \Rightarrow \boxed{I=5}$$


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