Difference between revisions of "2020 AIME I Problems/Problem 4"
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== Problem == | == Problem == | ||
+ | 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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-molocyxu | -molocyxu | ||
+ | |||
+ | ==Solution 2 (Official MAA)== | ||
+ | Suppose that <math>N</math> has the required property. Then there are positive integers <math>k</math> and <math>m</math> such that <math>N = 10^4m + 2020 = k\cdot m</math>. Thus <math>(k - 10^4)m = 2020</math>, which holds exactly when <math>m</math> is a positive divisor of <math>2020.</math> The number <math>2020 = 2^2\cdot 5\cdot 101</math> has <math>12</math> divisors: <math>1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010</math>, and <math>2020.</math> The requested sum is therefore the sum of the digits in these divisors plus <math>12</math> times the sum of the digits in <math>2020,</math> which is | ||
+ | <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>. | ||
+ | |||
+ | ==Video Solutions== | ||
+ | *https://youtu.be/5b9Nw4bQt_k | ||
+ | *https://youtu.be/djWzRC-jGYw | ||
==See Also== | ==See Also== |
Latest revision as of 14:02, 1 August 2022
Contents
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) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.