Let A = nN : n is a 3-digit number, B=9k+2 : kN and C=9k+l : kN for some l (0

Question

Let $A = \{n\in\mathbb{N} : n 	ext{ is a 3-digit number}\}$, $B=\{9k+2 : k\in\mathbb{N}\}$ and $C=\{9k+l : k\in\mathbb{N}\}$ for some $l$ ($0<l<9$). If the sum of all the elements of the set $A\cap(B\cup C)$ is $274	imes400$, then $l$ is equal to:

Answer: 5

Accepted Answer

3-digit numbers of the form $9k+2$: $101,109,\ldots,992$. Sum $=\dfrac{100}{2}(101+992)=100	imes\dfrac{1093}{2}=54650$.

For the total sum to equal $274	imes400=109600$, the remaining series (of form $9k+l$) must contribute $109600-54650=54950$.

Testing $l=5$: 3-digit numbers of the form $9k+5$ are $104,113,\ldots,995$. Sum $=\dfrac{100}{2}(104+995)=100	imes\dfrac{1099}{2}=54950$.  Matches exactly.

So $l=5$.

Solutions · 1