2020 AIME I Problems/Problem 4

Revision as of 00:46, 4 July 2021 by MRENTHUSIASM (talk | contribs) (Solution 1: Deleted extra line.)

Problem

Let $S$ be the set of positive integers $N$ with the property that the last four digits of $N$ are $2020,$ and when the last four digits are removed, the result is a divisor of $N.$ For example, $42,020$ is in $S$ because $4$ is a divisor of $42,020.$ Find the sum of all the digits of all the numbers in $S.$ For example, the number $42,020$ contributes $4+2+0+2+0=8$ to this total.

Solution 1

We note that any number in $S$ can be expressed as $a(10,000) + 2,020$ for some integer $a$. The problem requires that $a$ divides this number, and since we know $a$ divides $a(10,000)$, we need that $a$ divides 2020. Each number contributes the sum of the digits of $a$, as well as $2 + 0 + 2 +0 = 4$. Since $2020$ can be prime factorized as $2^2 \cdot 5 \cdot 101$, it has $(2+1)(1+1)(1+1) = 12$ factors. So if we sum all the digits of all possible $a$ values, and add $4 \cdot 12 = 48$, we obtain the answer.

Now we list out all factors of $2,020$, or all possible values of $a$. $1,2,4,5,10,20,101,202,404,505,1010,2020$. If we add up these digits, we get $45$, for a final answer of $45+48=\boxed{093}$.

-molocyxu

Solution 2 (Official MAA)

Suppose that $N$ has the required property. Then there are positive integers $k$ and $m$ such that $N = 10^4m + 2020 = k\cdot m$. Thus $(k - 10^4)m = 2020$, which holds exactly when $m$ is a positive divisor of $2020.$ The number $2020 = 2^2\cdot 5\cdot 101$ has $12$ divisors: $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010$, and $2020.$ The requested sum is therefore the sum of the digits in these divisors plus $12$ times the sum of the digits in $2020,$ which is \[(1+2+4+5+1+2+2+4+8+10+2+4)+12\cdot4 = 93.\]

Solution 3

Note that for all $N \in S$, $N$ can be written as $N=10000x+2020=20(500x+101)$ for some positive integer $x$. Because $N$ must be divisible by $x$, $\frac{20(500x+101)}{x}$ is an integer. We now let $x=ab$, where $a$ is a multiple of $20$. Then $\frac{20(500x+101)}{x}=(\frac{20}{a})( \frac{500x}{b}+\frac{101}{b})$. We know $\frac{20}{a}$ and $\frac{500x}{b}$ are integers, so for $N$ to be an integer, $\frac{101}{b}$ must be an integer. For this to happen, $b$ must be a multiple of $101$. $101$ is prime, so $b\in \left \{ 1, 101 \right \}$. Because $a$ is a multiple of $20$, $a \in \left \{ 1,2,4,5,10,20\right\}$. So $x \in \left\{1,2,4,5,10,20,101,202,404,505,1010,2020\right\}$. Be know that all $N$ end in $2020$, so the sum of the digits of each $N$ is the sum of the digits of each $x$ plus $2+0+2+0=4$. Hence the sum of all of the digits of the numbers in $S$ is $12 \cdot 4 +45=\boxed{093}$.

Video Solutions

See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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