Let

\begin{aligned} &\;\;\mathrm{A}=\{\mathrm{x}: \mathrm{x} \text { is } 3 \text { digit number }\}\\ &\begin{array}{l} \mathrm{B}=\{\mathrm{x}: \mathrm{x}=9 \mathrm{~K}+2, \mathrm{k} \in \mathrm{I}\} \\ \mathrm{C}=\{\mathrm{x}: \mathrm{x}=9 \mathrm{~K}+\ell, \mathrm{k} \in \mathrm{I}, \ell \in \mathrm{I}, 0<\ell<9\} \end{array} \end{aligned}

If the sum of all the element of the set A\cap \left ( B\cup C \right )is \; 274 \times \;400, then l is equals to:
Option: 1 1
Option: 2 3
Option: 3 5
Option: 4 7

Answers (1)

 

\begin{aligned} &\;\;\mathrm{A}=\{\mathrm{x}: \mathrm{x} \text { is } 3 \text { digit number }\}\\ &\begin{array}{l} \mathrm{B}=\{\mathrm{x}: \mathrm{x}=9 \mathrm{~K}+2, \mathrm{k} \in \mathrm{I}\} \\ \mathrm{C}=\{\mathrm{x}: \mathrm{x}=9 \mathrm{~K}+\ell, \mathrm{k} \in \mathrm{I}, \ell \in \mathrm{I}, 0<\ell<9\} \end{array} \end{aligned}

A \cap(B \cup C)=400\times274

Here A is the set of 3 digit number so we need to find the sum of 3-digit number of set B and set C.

3 digit number of the form 9 \mathrm{K}+2 are \{101,110, \ldots \ldots \ldots \ldots \ldots, 992\}

\Rightarrow S_{9k+2}= \frac{100}{2}(101+992)=50\times1093

Similarly'

3 digit number of the form 9 \mathrm{K}+l are \{99+l,108+l, \ldots \ldots \ldots \ldots \ldots, 990+l\}

\Rightarrow S_{9k+l}= \frac{100}{2}(99+l+990+l)=50\times(1089+2l)

S_{9k+l}+S_{9k+2}= 50\times(1089+2l)+50\times1093

400\times274=50\times(1089+2l)+50\times1093

8\times274=2182+2l

l+1091=1096

l=5

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