AMC 8 2016 Problem 11
AMC8 2016 年第 11 题
方程与不等式★★★☆☆
Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is 132.
有多少个两位数满足下面的性质：当把这个两位数的十位数字和个位数字调换位置所形成的新的数与原来的数相加，所得的和是 132．
- A.5
- B.7
- C.9
- D.11
- E.12
正确答案 · Correct Answer
B
解析 · Solution
Solution 1
We can see that the original number can be written as 10a+b, where represents the tens digit and represents the units digit. When this number is added to the number obtained by reversing its digits, which is 10b+a, the sum would be 11a+11b. From this, we can construct the equation 11a+11b=132, which simplifies to a+b=12. Since there are 7 pairs of such digits and , (3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3), the answer would be 7.
We can set the number as ab where a is the tens digit and b is the ones digit. So now the equation will be ab+ba=132. a+b has to have a remainder of 2 when divided by 10 so it will be 𝑎 + 𝑏 ≡ 2 ( mod 10 ). We also know a<10 and b<10. So a+b can either be 2 or 12. a+b cannot be 2 because then there will be only 3 numbers that work and that isn't in the answer choice. So a+b=12. To check this we can do ab+ba=132 which equals to (a+b)0+(a+b)=132 and since we said a+b=12 we get 120+12=132 which is true. So we have a+b=12 and a<10 and b<10.If a is 9 then b=3 and if a=3 then b=9 calculating how many pairs are in between you get 7. So the answer is 7.
Like similar solutions, we can say that 10 𝑎 + 𝑏 is a two digit number, and therefore we have 10 𝑎 + 𝑏 + 10 𝑏 + 𝑎 = 132 which implies 11 𝑎 + 11 𝑏 = 132 which implies 𝑎 + 𝑏 = 12. We then use stars and bars to find out that there are ( 13 1 ) = 13 different possible sums. However, we want to satisfy 0 ≤ 𝑎 ≤ 9 and 0 ≤ 𝑏 ≤ 9, therefore we see that the sums 0 + 12, 1 + 11, 2 + 10, and 12 + 0, 11 + 1, and 10 + 2 do not work, and our answer is 13 − 6 = 7..
The two digit number 10A+B + the reversed number 10B+A is equal to 11A+11B = 132. Dividing both sides by 11 gives A + B = 12. Since we can't have 1 + 11, 2 + 10 etc. meaning there are 7. solutions.
继续练习