AMC8 2007

AMC8 2007 · Q18

AMC8 2007 · Q18. It mainly tests Remainders & modular arithmetic, Digit properties (sum of digits, divisibility tests).

The product of the two 99-digit numbers 303,030,303, ..., 030,303 and 505,050,505, ..., 050,505 has thousands digit A and units digit B. What is the sum of A and B?

两个 99 位数 303,030,303, ..., 030,303 和 505,050,505, ..., 050,505 的乘积，千位数字为 A，个位数字为 B。A 和 B 的和是多少？

(A) 3 3

(B) 5 5

(D) 8 8

(E) 10 10

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) To find $A$ and $B$, it is sufficient to consider only $303 \cdot 505$, because 0 is in the thousands place in both factors. So $A = 3$ and $B = 5$, and the sum is $A + B = 3 + 5 = 8$.

（D）要找出 $A$ 和 $B$，只需考虑 $303 \cdot 505$，因为两个因数的千位上都是 0。因此 $A = 3$、$B = 5$，所以它们的和为 $A + B = 3 + 5 = 8$。

Topics