AMC 8 2014 Problem 21

AMC8 2014 年第 21 题

余数与同余★★★☆☆

The 7 -digit numbers

\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}

and

\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}

are each multiples of 3 . Which of the following could be the value of

C

七位数

\underline{7}\,\underline{4}\,\underline{A}\,\underline{5}\,\underline{2}\,\underline{B}\,\underline{1}

和

\underline{3}\,\underline{2}\,\underline{6}\,\underline{A}\,\underline{B}\,\underline{4}\,\underline{C}

都是 3 的倍数。那么

C

可能是下列哪个值？

正确答案 · Correct Answer

解析 · Solution

Solution 1

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. To be a multiple of 3, A + B has to be either 2 or 5 or 8... and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10... and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1, so 1 is our answer. Also, you could just look up the answer. It's much easier and totally isn't wrong in any way.:) Concept Solution() for a number to be divisible by 3, the sum of its digits must be divisible by 3. In the number 74A52B1, 7+4+A+5+2+B+1 must be divisible by 3. We get 19+A+B. 21 is the closest multiple of 3 meaning that A=2, B=0, order doesn't matter. Now we plug in those values for 326AB4C. It will be 3+2+6+2+0+4+C to get 17+C. The closest multiple of 3 is 18. So that means 17+C=18. Solving for C, our answer is 1 -TheNerdWhoIsNerdy. If

n

is divisible by 3, the sum of it's digits should also be divisible by 3. This means that

19+A+B (\mod 3) \equiv 0

A+B+1 (\mod 3) \equiv 0

. For equation 2,

15+A+B+C (\mod 3) \equiv 0

A+B+C (\mod 3) \equiv 0

. Logically, you can see the correlation between our first and second equations so we can make the assumption, C =1

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