Let A = x: x is a 3-digit number, B=x: x=9k+2, kZ, and C=x: x=9k+, kZ, 0

Question

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

Answer: 5

Accepted Answer

$B\cap A = \{101,110,\ldots,992\}$ (100 terms), $S(B)=\dfrac{100}{2}(101+992)=54650$.

Since the given total $274\times400=109600 \ne S(B)$, $\ell\ne2$, so $B\cap C=\varnothing$ and $S(B\cup C)=S(B)+S(C)$.

$$S(C) = 9\sum_{k=11}^{110}k + 110\ell = 9\cdot\frac{100}{2}(11+110)+100\ell = 54450+100\ell.$$

$$54650+54450+100\ell = 109600 \implies 100\ell=500 \implies \ell=5.$$

Solutions · 1