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考题21-22 2014 AMC 8

2018-08-04 重点归纳

AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8数学竞赛试题及答案吧：

Problem 21

The $7$ -digit numbers $AMC 8数学竞赛$ and $AMC数学竞赛网$ are each multiples of $3$ . Which of the following could be the value of $C$ ?

$AMC 8数学竞赛试题及答案$

Solution

The sum of a number's digits $\mod{3}$ is congruent to the number $\pmod{3}$ . $74A52B1 \mod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \mod{3}$ is also congruent to 0. $7+4+5+2+1 \equiv 1 \pmod{3}$ , so $A+B\equiv 2 \pmod{3}$ . As we know, $326AB4C\equiv 0 \pmod{3}$ , so $AMC 8数学竞赛试题及答案$ , and therefore $A+B+C\equiv 0 \pmod{3}$ . We can substitute 2 for $A+B$ , so $2+C\equiv 0 \pmod{3}$ , and therefore $C\equiv 1\pmod{3}$ . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $AMC数学竞赛网$ .

Problem 22

A $2$ -digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number?

$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad \textbf{(E) }9$

Solution

We can think of the number as $10a+b$ , where a and b are digits. Since the number is equal to the product of the digits ( $a\cdot b$ ) plus the sum of the digits ( $a+b$ ), we can say that $10a+b=a\cdot b+a+b$ . We can simplify this to $10a=a\cdot b+a$ , and factor to $(10)a=(b+1)a$ . Dividing by $a$ , we have that $b+1=10$ . Therefore, the units digit, $b$ , is $AMC 8数学竞赛试题及答案$