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Italian Mathematical Olympiad - February Round

Number theory Difficulty 4.6 AIME Find the answer Italy

Problem:

What is the sum of the positive divisors of 1800018000 whose decimal representation ends in 5050?

Pick one

Solution

Solution:

The answer is (D)(\mathbf{D}). We have 18000=24⋅32⋅5318000=2^{4} \cdot 3^{2} \cdot 5^{3}, so all divisors of 1800018000 are of the form 2a⋅3b⋅5c2^{a} \cdot 3^{b} \cdot 5^{c}, where a∈{0,1,2,3,4}a \in\{0,1,2,3,4\}, b∈{0,1,2}b \in\{0,1,2\} and c∈{0,1,2,3}c \in\{0,1,2,3\}. In order for the decimal representation of the divisor to end in 5050, the number must be divisible by 5050, hence a≥1a \geq 1 and c≥2c \geq 2. Note, moreover, that if a≥2a \geq 2 and c≥2c \geq 2, then the decimal representation of the divisor ends in the digits "00". It is therefore also necessary to require a=1a=1. Hence the required sum is given by
∑b=02∑c=232⋅3b⋅5c=2⋅(∑b=023b)⋅(∑c=235c)=2⋅13⋅150=3900. \sum_{b=0}^{2} \sum_{c=2}^{3} 2 \cdot 3^{b} \cdot 5^{c} = 2 \cdot \left(\sum_{b=0}^{2} 3^{b}\right) \cdot \left(\sum_{c=2}^{3} 5^{c}\right) = 2 \cdot 13 \cdot 150 = 3900.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.