AMC 8 2019 Problem 13
AMC8 2019 年第 13 题
整除与因数★★★☆☆
A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of ?
回环数是指从左向右读和从右向左读，读数是一样的数（例如，12321 是个回环数）。N 是满足以下条件的最小三位整数：它不是个回环数，并且它是 3 个不同的两位回环数的和。问 N 的各个位上的数字之和是多少？
- A.2
- B.3
- C.4
- D.5
- E.6
正确答案 · Correct Answer
A
解析 · Solution
Solution 1
Note that the only positive 2-digit palindromes are multiples of 11, namely . Since is the sum of 2-digit palindromes, is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so N=110 is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as 110=77+22+11. Then, N = 110, and the sum of the digits of is 1+1+0 = 2. There are other sets of 2-digit numbers that satisfy this rule. Some of them are 110 = 11+33+66 and 110 = 22+33+55 We already know that two-digit palindromes can only be two-digit multiples of 11; which are: 11, 22, 33, 44, 55, 66, 77, 88, and 99. Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then, we start counting. Aha! This multiple of 11, 110, not only isn't a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of 's digits is 1+1+0 = 2. Thank you to the writer of for inspiring me to create this! EarthSaver 15:13, 11 June 2021 (EDT) As stated above, two-digit palindromes can only be two-digit multiples of 11. We can see that if we add anything that are multiples of 11 together, we will again get a multiple of 11. For instance, 11+22=33. Since we know this fact and we are finding the smallest value possible, we can start with the first three-digit multiple of 11 which is 110. Since this is not a palindrome and can be the sum of 3 two-digit palindromes (see above solutions for more details), 110 fits the bill. We can see that the sum of 110 's digits is 1+1+0 = 2. Two digit palindromes are two digit multiples of 11. When we take 33×3, we get 99, therefore 4 numbers that work for this would be 33, 33, 33 and 11. To convert this into three numbers, we can subtract 11 from the first 33, then add both 11 s to the last 33 to get the set 22, 33 and 55. 22+33+55 = 110 and 1+1+0 = 2. ~
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