2013 AMC 10B Problem 18

Attempt Problem 18 of the 2013 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 10B solutions, or check the answer key.

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18.

The number 20132013 has the property that its units digit is the sum of its other digits, that is 2+0+1=3.2+0+1=3. How many integers less than 20132013 but greater than 10001000 have this property?

33 33

34 34

45 45

46 46

58 58

Answer: D
Concepts:digitscaseworktriangular number
Difficulty rating: 1420
Solution:

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

Now, we can case on the 11st digit.

If it is 1, then the sum of the 22nd and 33rd digit must be less than or equal to 8.8. For each possible sum s,s, there are s+1s+1 ways to choose the other numbers as the 2nd number can be anywhere from 00 to s.s.

Thus, the total is the 9th9^{th} triangular number: 9102=45.\dfrac{9\cdot 10}2 =45.

If it is 2, then the only way we can get a number that works less than 20132013 is 2002,2002, making a total of 4646 cases.

Thus, the correct answer is D .

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