Difference between revisions of "2020 AIME I Problems/Problem 4"
MRENTHUSIASM (talk | contribs) m (→Solution 1: Deleted extra line.) |
(→Solution 3) |
||
Line 16: | Line 16: | ||
==Solution 3== | ==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 | + | 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== | ==Video Solutions== |
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) | ||
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.