AMC10 2013 B

AMC10 2013 B · Q18

AMC10 2013 B · Q18. It mainly tests Basic counting (rules of product/sum), Digit properties (sum of digits, divisibility tests).

The number 2013 has the property that its units digit is the sum of its other digits, that is 2 + 0 + 1 = 3. How many integers less than 2013 but greater than 1000 share this property?

数 2013 具有其个位数字等于其他数字之和的性质，即 $2+0+1=3$。有多少大于 1000 但小于 2013 的整数具有此性质？

(A) 33 33

(B) 34 34

(D) 46 46

(E) 58 58

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): First note that the only number between 2000 and 2013 that shares this property is 2002. Consider now the numbers in the range 1001 to 1999. There is exactly 1 number, 1001, that shares the property when the units digits is 1. There are exactly 2 numbers, 1102 and 1012, when the units digit is 2; exactly 3 numbers, 1203, 1113, and 1023, when the units digits is 3, and so on. Because the thousands digit is always 1, when the units digit is $n$, for $1 \le n \le 9$, the sum of the hundreds and tens digits must be $n-1$. There are exactly $n$ ways for this to occur. Hence there are exactly $$1+(1+2+\cdots+9)=1+\frac{9\cdot10}{2}=1+45=46$$ numbers that share this property.

答案（D）：首先注意到，在 2000 到 2013 之间，唯一具有该性质的数字是 2002。现在考虑 1001 到 1999 范围内的数。当个位数字为 1 时，恰好有 1 个数 1001 具有该性质。当个位数字为 2 时，恰好有 2 个数 1102 和 1012；当个位数字为 3 时，恰好有 3 个数 1203、1113 和 1023；以此类推。由于千位数字总是 1，当个位数字为 $n$（$1 \le n \le 9$）时，百位与十位数字之和必须为 $n-1$。满足这一条件的方式恰好有 $n$ 种。因此，具有该性质的数字总共有 $$1+(1+2+\cdots+9)=1+\frac{9\cdot10}{2}=1+45=46$$ 个。

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