Difference between revisions of "2020 AIME I Problems/Problem 4"
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Let <math>S</math> be the set of positive integers <math>N</math> with the property that the last four digits of <math>N</math> are <math>2020,</math> and when the last four digits are removed, the result is a divisor of <math>N.</math> For example, <math>42,020</math> is in <math>S</math> because <math>4</math> is a divisor of <math>42,020.</math> Find the sum of all the digits of all the numbers in <math>S.</math> For example, the number <math>42,020</math> contributes <math>4+2+0+2+0=8</math> to this total. | Let <math>S</math> be the set of positive integers <math>N</math> with the property that the last four digits of <math>N</math> are <math>2020,</math> and when the last four digits are removed, the result is a divisor of <math>N.</math> For example, <math>42,020</math> is in <math>S</math> because <math>4</math> is a divisor of <math>42,020.</math> Find the sum of all the digits of all the numbers in <math>S.</math> For example, the number <math>42,020</math> contributes <math>4+2+0+2+0=8</math> to this total. | ||
− | == Solution == | + | == Solution 1 == |
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. | 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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==Solution 2 (Official MAA)== | ==Solution 2 (Official MAA)== | ||
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<cmath>(1+2+4+5+1+2+2+4+8+10+2+4)+12\cdot4 = 93.</cmath> | <cmath>(1+2+4+5+1+2+2+4+8+10+2+4)+12\cdot4 = 93.</cmath> | ||
− | + | ==Solution 3== | |
+ | Note that for all <math>N \in S</math>, <math>N</math> can be written as <math>N=10000x+2020=20(500x+101)</math> for some positive integer <math>x</math>. Because <math>N</math> must be divisible by <math>x</math>, <math>\frac{20(500x+101)}{x}</math> is an integer. We now let <math>x=ab</math>, where <math>a</math> is a divisor of <math>20</math>. Then <math>\frac{20(500x+101)}{x}=(\frac{20}{a})( \frac{500x}{b}+\frac{101}{b})</math>. We know <math>\frac{20}{a}</math> and <math>\frac{500x}{b}</math> are integers, so for <math>N</math> to be an integer, <math>\frac{101}{b}</math> must be an integer. For this to happen, <math>b</math> must be a divisor of <math>101</math>. <math>101</math> is prime, so <math>b\in \left \{ 1, 101 \right \}</math>. Because <math>a</math> is a divisor of <math>20</math>, <math>a \in \left \{ 1,2,4,5,10,20\right\}</math>. So <math>x \in \left\{1,2,4,5,10,20,101,202,404,505,1010,2020\right\}</math>. Be know that all <math>N</math> end in <math>2020</math>, so the sum of the digits of each <math>N</math> is the sum of the digits of each <math>x</math> plus <math>2+0+2+0=4</math>. Hence the sum of all of the digits of the numbers in <math>S</math> is <math>12 \cdot 4 +45=\boxed{093}</math>. | ||
− | https://youtu.be/djWzRC-jGYw | + | ==Video Solutions== |
+ | *https://youtu.be/5b9Nw4bQt_k | ||
+ | *https://youtu.be/djWzRC-jGYw | ||
==See Also== | ==See Also== |
Latest revision as of 13:02, 1 August 2022
Contents
[hide]Problem
Let be the set of positive integers with the property that the last four digits of are and when the last four digits are removed, the result is a divisor of For example, is in because is a divisor of Find the sum of all the digits of all the numbers in For example, the number contributes to this total.
Solution 1
We note that any number in can be expressed as for some integer . The problem requires that divides this number, and since we know divides , we need that divides 2020. Each number contributes the sum of the digits of , as well as . Since can be prime factorized as , it has factors. So if we sum all the digits of all possible values, and add , we obtain the answer.
Now we list out all factors of , or all possible values of . . If we add up these digits, we get , for a final answer of .
-molocyxu
Solution 2 (Official MAA)
Suppose that has the required property. Then there are positive integers and such that . Thus , which holds exactly when is a positive divisor of The number has divisors: , and The requested sum is therefore the sum of the digits in these divisors plus times the sum of the digits in which is
Solution 3
Note that for all , can be written as for some positive integer . Because must be divisible by , is an integer. We now let , where is a divisor of . Then . We know and are integers, so for to be an integer, must be an integer. For this to happen, must be a divisor of . is prime, so . Because is a divisor of , . So . Be know that all end in , so the sum of the digits of each is the sum of the digits of each plus . Hence the sum of all of the digits of the numbers in is .
Video Solutions
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
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