AMC8 2019

AMC8 2019 · Q13

AMC8 2019 · Q13. It mainly tests Basic counting (rules of product/sum), Primes & prime factorization.

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 $N$ 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 $N$?

回文数是一个从左到右读或从右到左读值相同的数。（例如，12321是一个回文数。）设$N$为最小的一个不是回文数但等于三个不同两位回文数之和的三位整数。$N$的各位数字之和是多少？

(A) 2 2

(B) 3 3

(D) 5 5

(E) 6 6

Answer

Correct choice: (A)

正确答案：（A）

Solution

Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \ldots, 99$. Since $N$ is the sum of 2-digit palindromes, $N$ 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 $N$ is $1+1+0 = \boxed{\textbf{(A) }2}$.

注意，唯一正的两位的回文数是11的倍数，即$11, 22, \ldots, 99$。由于$N$是两位回文数之和，$N$必然是11的倍数。最小不是回文数的三位11的倍数是110，所以$N=110$是一个候选解。我们必须检查110能否写成三个不同两位回文数之和；$110=77+22+11$满足。于是，$N = 110$，$N$的各位数字之和是$1+1+0 = \boxed{\textbf{(A) }2}$。

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