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Difference between revisions of "2020 AIME I Problems/Problem 4"
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== Solution ==
 
== Solution ==
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We note that any number in <math>S</math> can be expressed as <math>a(10,000) + 2,020</math> for some integer <math>a</math>. The problem requires that <math>a</math> divides this number, and since we know <math>a</math> divides <math>a(10,000)</math>, we need that <math>a</math> divides 2020. Each number contributes the sum of the digits of <math>a</math>, as well as <math>2 + 0 + 2 +0 = 4</math>. Since <math>2020</math> can be prime factorized as <math>2^2 \cdot 5 \cdot 101</math>, it has <math>(2+1)(1+1)(1+1) = 12</math> factors. So if we sum all the digits of all possible <math>a</math> values, and add <math>4 \cdot 12 = 48</math>, we obtain the answer.
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Now we list out all factors of <math>2,020</math>, or all possible values of <math>a</math>. <math>1,2,4,5,10,20,101,202,404,505,1010,2020</math>. If we add up these digits, we get <math>45</math>, for a final answer of <math>45+48=\boxed{093}</math>.
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-molocyxu
  
 
==See Also==
 
==See Also==

Revision as of 16:12, 12 March 2020

Note: Please do not post problems here until after the AIME.

Problem

Solution

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

See Also

2020 AIME I (Problems • Answer Key • Resources)
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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