Question

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 $1000$ have this property?

$33$

$34$

$45$

$46$

$58$

Solution:

Given the first three numbers, if their sum is less than or equal to $9,$ it creates one number with the property.

Now, we can case on the $1$ st digit.

If it is 1, then the sum of the $2$ nd and $3$ rd digit must be less than or equal to $8.$ For each possible sum $s,$ there are $s+1$ ways to choose the other numbers as the 2nd number can be anywhere from $0$ to $s.$

Thus, the total is the $9^{th}$ triangular number: $\dfrac{9\cdot 10}2 =45.$

If it is 2, then the only way we can get a number that works less than $2013$ is $2002,$ making a total of $46$ cases.

Thus, the correct answer is D .

2013 AMC 10B Problem 18

Solution:

Problem 18 in Other Years