AMC 8 2016 Problem 5
AMC8 2016 年第 5 题
余数与同余★★☆☆☆
The number N is a two-digit number. • When N is divided by 9, the remainder is 1. • When N is divided by 10, the remainder is 3. What is the remainder when N is divided by 11?
N 是个两位数。
·当 N 除以 9，余数为 1． ·当 N 除以 10，余数为 3． 当 N 除以 11 时，余数是多少？
- A.0
- B.2
- C.4
- D.5
- E.7
正确答案 · Correct Answer
E
解析 · Solution
Solution 1
From the second bullet point, we know that the second digit must be 3, for a number divisible by 10 ends in zero. Since there is a remainder of 1 when is divided by 9, the multiple of 9 must end in a 2 for it to have the desired remainder We now look for this one: 9(1)=9\9(2)=18\9(3)=27\9(4)=36\9(5)=45\9(6)=54\9(7)=63\9(8)=72 The number 72+1=73 satisfies both conditions. We subtract the biggest multiple of 11 less than 73 to get the remainder. Thus, 73-11(6)=73-66=7.
We know that the number has to be one more than a multiple of 9, because of the remainder of one, and the number has to be 3 more than a multiple of 10, which means that it has to end in a 3. Now, if we just list the first few multiples of 9 adding one to the number we get: 10, 19, 28, 37, 46, 55, 64, 73, 82, 91. As we can see from these numbers, the only one that has a three in the units place is 73, thus we divide 73 by 11, getting 6R7, hence, 7. -fn106068 We could also remember that, for a two-digit number to be divisible by 9, the sum of its digits has to be equal to 9. Since the number is one more than a multiple of 9, the multiple we are looking for has a ones digit of 2, and therefore a tens digit of 9-2 = 7, and then we could proceed as above. -vaisri
We can set up a system of modular congruences n = 1 (mod 9) n = 3 (mod 10) 9x + 1 = 3 (mod 10) -1 -1 9x = 2(mod 10) +70 9x = 72(mod 10) /9 /9 x = 8 (mod 10) 10f + 8 9(10f+8)+1= n 90f + 72 + 1 = n 90f + 73 = n n = 73 (mod 90) = 6 r7 - timi821
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