AMC10 2003 B

AMC10 2003 B · Q25

AMC10 2003 B · Q25. It mainly tests Basic counting (rules of product/sum), Digit properties (sum of digits, divisibility tests).

How many distinct four-digit numbers are divisible by 3 and have 23 as their last two digits?

有多少个不同的四位数能被3整除且末两位数是23？

(A) 27 27

(B) 30 30

(D) 81 81

(E) 90 90

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) A number is divisible by 3 if and only if the sum of its digits is divisible by 3. So a four-digit number $ab23$ is divisible by 3 if and only if the two-digit number $ab$ leaves a remainder of 1 when divided by 3. There are 90 two-digit numbers, of which $90/3 = 30$ leave a remainder of 1 when divided by 3.

（B）一个数能被 3 整除，当且仅当它的各位数字之和能被 3 整除。因此，四位数 $ab23$ 能被 3 整除，当且仅当两位数 $ab$ 除以 3 的余数为 1。两位数共有 90 个，其中有 $90/3 = 30$ 个除以 3 的余数为 1。

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