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美国数学竞赛AMC8真题2014年 21-22

2018-09-20 重点归纳

美国数学竞赛AMC8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年级的考生加入到美国数学竞赛AMC8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美国数学竞赛AMC8的官方真题以及官方解答吧：

Problem 21

The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $amc数学竞赛8$ are each multiples of $3$ . Which of the following could be the value of $C$ ?

$美国数学竞赛AMC8$

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. $amc真题$ , so $A+B\equiv 2 \pmod{3}$ . As we know, $326AB4C\equiv 0 \pmod{3}$ , so $3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}$ , and therefore $A+B+C\equiv 0 \pmod{3}$ . We can substitute 2 for $A+B$ , so $amc 真题$ , 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 $\boxed{\textbf{(A) }1}$ .

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?

$美国数学竞赛amc8$

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 $美国数学竞赛amc8$ , and factor to $(10)a=(b+1)a$ . Dividing by $a$ , we have that $美国数学竞赛amc8真题及答案解析$ . Therefore, the units digit, $b$ , is $\boxed{\textbf{(E) }9}$