AMC8 1993

AMC8 1993 · Q19

AMC8 1993 · Q19. It mainly tests Arithmetic sequences basics.

(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193) =

$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193) =

(A) 167,400 167,400

(B) 172,050 172,050

(D) 199,300 199,300

(E) 362,142 362,142

Answer

Correct choice: (A)

正确答案：（A）

Solution

English (LaTeX-ready): Answer (A): Each number in the first set of numbers is 1800 more than the corresponding number in the second set: $\begin{array}{cccc} 1901, & 1902, & 1903, & \ldots,\ 1993 \\ -101, & -102, & -103, & \ldots,\ -193 \\ 1800, & 1800, & 1800, & \ldots,\ 1800 \end{array}$ Thus the sum of the first set of numbers $93 \times 1800 = 167{,}400$ more than the sum of the second set.

中文（翻译）：答案（A）：第一组中的每个数都比第二组中对应的数大 $1800$： $\begin{array}{cccc} 1901, & 1902, & 1903, & \ldots,\ 1993 \\ -101, & -102, & -103, & \ldots,\ -193 \\ 1800, & 1800, & 1800, & \ldots,\ 1800 \end{array}$ 因此，第一组数的和比第二组数的和多 $93 \times 1800 = 167{,}400$。

Topics

AR.012 Arithmetic sequences basics