Difference between revisions of "2019 AIME II Problems/Problem 9"
(→Solution 2) |
|||
(2 intermediate revisions by one other user not shown) | |||
Line 16: | Line 16: | ||
==Solution 2== | ==Solution 2== | ||
− | For <math>n</math> to have exactly <math>20</math> positive divisors, <math>n</math> can only take on certain prime factorization forms: namely, <math>p^{19}, p^9q, p^4q^3, p^4qr</math>. No number that is a multiple of <math>20</math> can be expressed in the first form, and the only integer divisible by <math>20</math> that has the second form is <math>2^{9}5</math>, which is greater than <math>2019</math>. | + | For <math>n</math> to have exactly <math>20</math> positive divisors, <math>n</math> can only take on certain prime factorization forms: namely, <math>p^{19}, p^9q, p^4q^3, p^4qr</math> where <math>p,q,r</math> are primes. No number that is a multiple of <math>20</math> can be expressed in the first form because 20 has ''two'' primes in its prime factorization, while the first form has only ''one'', and the only integer divisible by <math>20</math> that has the second form is <math>2^{9}5</math>, which is 2560, greater than <math>2019</math>. |
For the third form, the only <math>20</math>-pretty numbers are <math>2^45^3=2000</math> and <math>2^35^4=5000</math>, and only <math>2000</math> is small enough. | For the third form, the only <math>20</math>-pretty numbers are <math>2^45^3=2000</math> and <math>2^35^4=5000</math>, and only <math>2000</math> is small enough. | ||
Line 23: | Line 23: | ||
Thus, <math>\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}</math>. | Thus, <math>\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}</math>. | ||
+ | |||
+ | Rephrased for clarity by Afly | ||
==Solution 3== | ==Solution 3== | ||
− | The divisors of <math>20</math> are <math>{1,2,4,5,10,20}</math>. <math> | + | The divisors of <math>20</math> are <math>{1,2,4,5,10,20}</math>. <math>v_2(n)</math> must be <math>\ge 2</math> because <math>20=2^2 \times 5</math>. This means that <math>v_2(n)</math> can be exactly <math>3</math> or <math>4</math>. Because greater exponents of 2 (like <math>2^9\cdot5</math>) gives numbers greater than 2019, so we just have the cases below. |
− | 1. <math> | + | 1. <math>v_2(n) = 3</math>. Then <math>\frac{20}{4}=5=5\times 1</math>. The smallest is <math>2^3*5^4</math> which is <math>> 2019</math>. Hence there are no solution in this case. |
− | 2. <math> | + | 2. <math>v_2(n)=4</math>. Then <math>\frac{20}{5}=4 = 4\times 1 = 2\times 2</math>. |
The <math>4\times 1</math> case gives one solution, <math>2^4 \times 5^3 = 2000</math>. | The <math>4\times 1</math> case gives one solution, <math>2^4 \times 5^3 = 2000</math>. | ||
The <math>2\times 2</math> case gives <math>2^4\times 5 \times (3+7+11+13+17+19+23)</math>.Using any prime greater than <math>23</math> will make <math>n</math> greater than <math>2019</math>. | The <math>2\times 2</math> case gives <math>2^4\times 5 \times (3+7+11+13+17+19+23)</math>.Using any prime greater than <math>23</math> will make <math>n</math> greater than <math>2019</math>. |
Latest revision as of 14:46, 4 January 2024
Problem
Call a positive integer
-pretty if
has exactly
positive divisors and
is divisible by
. For example,
is
-pretty. Let
be the sum of positive integers less than
that are
-pretty. Find
.
Solution 1
Every 20-pretty integer can be written in form , where
,
,
, and
, where
is the number of divisors of
. Thus, we have
, using the fact that the divisor function is multiplicative. As
must be a divisor of 20, there are not many cases to check.
If , then
. But this leads to no solutions, as
gives
.
If , then
or
. The first case gives
where
is a prime other than 2 or 5. Thus we have
. The sum of all such
is
. In the second case
and
, and there is one solution
.
If , then
, but this gives
. No other values for
work.
Then we have .
-scrabbler94
Solution 2
For to have exactly
positive divisors,
can only take on certain prime factorization forms: namely,
where
are primes. No number that is a multiple of
can be expressed in the first form because 20 has two primes in its prime factorization, while the first form has only one, and the only integer divisible by
that has the second form is
, which is 2560, greater than
.
For the third form, the only -pretty numbers are
and
, and only
is small enough.
For the fourth form, any number of the form where
is a prime other than
or
will satisfy the
-pretty requirement. Since
,
. Therefore,
can take on
or
.
Thus, .
Rephrased for clarity by Afly
Solution 3
The divisors of are
.
must be
because
. This means that
can be exactly
or
. Because greater exponents of 2 (like
) gives numbers greater than 2019, so we just have the cases below.
1. . Then
. The smallest is
which is
. Hence there are no solution in this case.
2. . Then
.
The
case gives one solution,
.
The
case gives
.Using any prime greater than
will make
greater than
.
The answer is .
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.