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==Problem== | ==Problem== | ||
− | If the positive integer <math> | + | If the positive integer <math>n</math> has positive integer divisors <math>a</math> and <math>b</math> with <math>n = ab</math>, then <math>a</math> and <math>b</math> are said to be <math>\textit{complementary}</math> divisors of <math>n</math>. Suppose that <math>N</math> is a positive integer that has one complementary pair of divisors that differ by <math>20</math> and another pair of complementary divisors that differ by <math>23</math>. What is the sum of the digits of <math>N</math>? |
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math> | <math>\textbf{(A) } 9 \qquad \textbf{(B) } 13\qquad \textbf{(C) } 15 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math> | ||
== Solution 1 == | == Solution 1 == | ||
− | Consider positive <math>a, b</math> with a difference of <math>20</math>. Suppose <math>b = a-20</math>. Then, we have | + | Consider positive integers <math>a, b</math> with a difference of <math>20</math>. Suppose <math>b = a-20</math>. Then, we have <math>(a)(a-20) = n</math>. If there is another pair of two integers that multiply to <math>n</math> but have a difference of 23, one integer must be greater than <math>a</math>, and the other must be smaller than <math>a-20</math>. We can create two cases and set both equal. We have <math>(a)(a-20) = (a+1)(a-22)</math>, and <math>(a)(a-20) = (a+2)(a-21)</math> (under the requirement that one of the variables in the second case must be smaller than <math>a-20</math>). Starting with the first case, we have <math>a^2-20a = a^2-21a-22</math>,or <math>0=-a-22</math>, which gives <math>a=-22</math>, which is not possible. The other case is <math>a^2-20a = a^2-19a-42</math>, so <math>a=42</math>. Thus, our product is <math>(42)(22) = (44)(21)</math>, so <math>c = 924</math>. Adding the digits, we have <math>9+2+4 = \boxed{\textbf{(C) } 15}</math>. |
− | -Sepehr2010 | + | |
+ | -Sepehr2010, minor edits by the_eaglercraft_grinder | ||
==Solution 2 == | ==Solution 2 == | ||
Line 27: | Line 28: | ||
x^2+20x&=y^2+23y\\ | x^2+20x&=y^2+23y\\ | ||
4x^2+4\cdot20x &= 4y^2+4\cdot23y\\ | 4x^2+4\cdot20x &= 4y^2+4\cdot23y\\ | ||
− | 4x^2+4\cdot20x+20^2-20^2 &= 4y^2+4\cdot23y+23^ | + | 4x^2+4\cdot20x+20^2-20^2 &= 4y^2+4\cdot23y+23^2-23^2\\ |
(2x+20)^2-20^2 &= (2y+23)^2-23^2\\ | (2x+20)^2-20^2 &= (2y+23)^2-23^2\\ | ||
23^2-20^2 &= (2y+23)^2-(2x+20)^2\\ | 23^2-20^2 &= (2y+23)^2-(2x+20)^2\\ | ||
Line 37: | Line 38: | ||
<cmath>\begin{align} | <cmath>\begin{align} | ||
129 or 43 &= (2y+2x+43)\\ | 129 or 43 &= (2y+2x+43)\\ | ||
− | 1 or 3&= 2y-2x+3\\ | + | 1 or 3 &= 2y-2x+3\\ |
\end{align}</cmath> | \end{align}</cmath> | ||
43 & 1 yields (0,0) which is not what we want. | 43 & 1 yields (0,0) which is not what we want. | ||
Line 58: | Line 59: | ||
== Solution 4 == | == Solution 4 == | ||
− | Say one factorization is <math>n(n+23).</math> The two cases for the other factorization are <math>(n+1)(n+21)</math> and <math>(n+2)(n+22).</math> We know it must be the first because of AM-GM intuition: lesser factors are closer together. Thus, <math>n(n+23)=(n+1)(n+21)</math> and we find that <math>n=21, | + | Say one factorization is <math>n(n+23).</math> The two cases for the other factorization are <math>(n+1)(n+21)</math> and <math>(n+2)(n+22).</math> We know it must be the first because of AM-GM intuition: lesser factors of a number are closer together than larger factors of a number. (We can also try both and see which works.) Thus, <math>n(n+23)=(n+1)(n+21)</math> and we find that <math>n=21,N=924</math> meaning the answer is <math>\boxed{\textbf{(C) }15}.</math> |
~DouDragon | ~DouDragon | ||
Line 64: | Line 65: | ||
== Solution 5 == | == Solution 5 == | ||
− | Since we are given that some pairs of divisors differ by 20 and 23 and we can let the pair. We also know the product of both the complementary divisors give the same number so <math>(x-10)(x+10)=(y-\frac{23}{2})(y+\frac{23}{2})</math> . | + | Since we are given that some pairs of divisors differ by 20 and 23 and we can let the pair be <math>(x-10)</math> and <math>(x+10)</math> as well as <math>(y-\frac{23}{2})</math> and <math>(y+\frac{23}{2})</math> . We also know the product of both the complementary divisors give the same number so <math>(x-10)(x+10)=(y-\frac{23}{2})(y+\frac{23}{2})</math> . |
Now we let <math>y=\frac{a}{2}</math>. Then we substitute and get <math>x^2-100=\frac{(a^2-529)}{4}</math>. Finally we multiply by 4 and get <math>4x^2-a^2=-129, a^2-4x^2=129</math>. | Now we let <math>y=\frac{a}{2}</math>. Then we substitute and get <math>x^2-100=\frac{(a^2-529)}{4}</math>. Finally we multiply by 4 and get <math>4x^2-a^2=-129, a^2-4x^2=129</math>. | ||
Then we use differences of squares and get <math>a</math>+<math>2x</math>=129, <math>a</math>-<math>2x</math>=1. We finish by getting <math>a=</math>65 and <math>x=32</math>. So <math>(42)(22) = 924</math> Adding the digits, we have <math>9+2+4 = \boxed{\textbf{(C) } 15}</math>. | Then we use differences of squares and get <math>a</math>+<math>2x</math>=129, <math>a</math>-<math>2x</math>=1. We finish by getting <math>a=</math>65 and <math>x=32</math>. So <math>(42)(22) = 924</math> Adding the digits, we have <math>9+2+4 = \boxed{\textbf{(C) } 15}</math>. | ||
− | - | + | |
+ | |||
+ | ~averageguy | ||
+ | |||
+ | |||
+ | Nunber sense note: To avoid tedious multiplication of 2-digit numbers, observe that <math>n = (42)(22) = (6)(7)(2)(11)</math>, and <math>(6)(7)(2) = 84</math>, and the sum of the digits of <math>11</math> is <math>2</math>, so the sum of the digits of <math>n</math> is equivalent to <math>(8+4)(2) \equiv 24 \equiv 15 \pmod 9</math>. The only equivalent answer choice is <math>\boxed{15}</math>. ~oinava | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | <math>N</math> can be written <math>N = \left( a - 10 \right) \left( a + 10 \right)</math> with a positive integer <math>a > 10</math> and <math>N = \left( \frac{2b + 1}{2} - \frac{23}{2} \right) \left( \frac{2b + 1}{2} + \frac{23}{2} \right)</math> with a positive integer <math>b > 11</math>. | ||
+ | |||
+ | The above equations can be reorganized as | ||
+ | <cmath> | ||
+ | \[ | ||
+ | \left( 2b + 1 + 2 a \right) \left( 2 b + 1 - 2 a \right) | ||
+ | = 43 \cdot 3 . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | The only solution is <math>2b + 1 + 2a = 129</math> and <math>2b + 1 - 2a = 1</math>. | ||
+ | Thus, <math>a = b = 32</math>. | ||
+ | Therefore, <math>N = 924</math>. | ||
+ | So the sum of the digits of <math>N</math> is <math>9 + 2 + 4 = \boxed{\textbf{(C)}~15}</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 7== | ||
+ | We can write <math>N</math> as <math>a(a+20)</math> or <math>b(b+23)</math> where <math>a</math> and <math>b</math> are divisors of <math>N.</math> Since <math>a(a+20) = b(b+23),</math> we know that <math>a^2 + 20a - b^2 - 23b = 0</math>, and we can view this as a quadratic in <math>a.</math> | ||
+ | |||
+ | Since the solution for <math>a</math> must be an integer, the discriminant for this quadratic must be a perfect square and therefore <math>20^2 - 4(-b^2 - 23b) = (2c)^2 = 400 + 4b^2 + 92b</math> so <math>b^2 + 23b -c^2 + 100 = 0.</math> | ||
+ | |||
+ | Since the discriminant of this quadratic in <math>b</math> must also be a perfect square we know that <math>23^2 - 4(-c^2+100) = d^2</math> which we can simplify as <math>d^2 - 4c^2 = (d-2c)(d+2c) = 129.</math> Since they are both positive integers <math>d - 2c</math> and <math>d + 2c</math> are factors of <math>129 = 3 \cdot 43</math> so <math>d - 2c = 1</math> and <math>d + 2c = 129</math> or <math>d - 2c = 3</math> and <math>d - 2c = 43.</math> | ||
+ | |||
+ | These systems of equations give us <math>(c,d) = (32,65)</math> and <math>(c,d) = (10,23)</math> respectively, if we plug our values for <math>c</math> into the equation for <math>b</math> we get <math>b^2 + 23b - 924 = 0</math> and <math>b^2 + 23b = 0</math> respectively. The first equation gives us <math>b = 21</math> or <math>b = -44</math> and the second gives us <math>b = 0</math> or <math>b = -23</math>, since <math>b</math> is positive we know that <math>b = 21</math> and <math>N = (21)(21 + 23) = 924</math>, therefore the sum of the digits of <math>N</math> is <math>9 + 2 + 4 = \boxed{\textbf{(C) 15}}.</math> | ||
+ | |||
+ | ~SailS | ||
+ | |||
+ | ==Solution 8 (Trial and Error)== | ||
+ | Consider the numbers of the form <math>a(a+20)</math>. Since <math>b(b+23)</math> is always even, <math>a</math> is even. Thus, for <math>a \ge 2</math>, we calculate <math>a(a+20)</math> for even values of <math>a</math>. Then, we check if it can also be represented as a product of numbers that differ by <math>23</math>. Checking, we see that <math>22 \cdot 42 = 21 \cdot 44 = 924</math> works. Thus, the answer is <math>9 + 2 + 4 = \boxed{\textbf{(C) 15}}</math> | ||
+ | |||
+ | ~andliu766 | ||
+ | |||
+ | ==Solution 9== | ||
+ | |||
+ | <math>n(n+20)=m(m+23) \Longrightarrow n^2+20n=m^2+23m \Longrightarrow n^2-m^2+20n-20m=3m</math>. Factoring, <math>(n+m)(n-m)+20(n-m)=3m \Longrightarrow (n-m)(n+m+20)=3m</math>. Let <math>n-m=a>0</math> because clearly <math>n>m</math>. Then. <math>a(2m+20+a)=3m</math>. Note that since <math>20+a>0</math>, if <math>a\geq2</math>, then the equation is <math>4m+a(20+a)>3m</math>, so <math>a-1</math>. Plugging this back, we get <math>2m+21=3m \Longrightarrow m=21</math> and <math>n=22</math>. Now we find <math>N</math> as <math>22*42=924</math> so the answer is <math>15</math>. | ||
+ | |||
+ | -Magnetoninja | ||
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == | ||
https://youtu.be/D_T24PrVk18 | https://youtu.be/D_T24PrVk18 | ||
+ | |||
+ | ==Video Solution by epicbird08== | ||
+ | https://youtu.be/HrZ3fia7g2A | ||
+ | |||
+ | ~EpicBird08 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/J9VAVT22L40 | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2023|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:04, 24 July 2024
Contents
Problem
If the positive integer has positive integer divisors
and
with
, then
and
are said to be
divisors of
. Suppose that
is a positive integer that has one complementary pair of divisors that differ by
and another pair of complementary divisors that differ by
. What is the sum of the digits of
?
Solution 1
Consider positive integers with a difference of
. Suppose
. Then, we have
. If there is another pair of two integers that multiply to
but have a difference of 23, one integer must be greater than
, and the other must be smaller than
. We can create two cases and set both equal. We have
, and
(under the requirement that one of the variables in the second case must be smaller than
). Starting with the first case, we have
,or
, which gives
, which is not possible. The other case is
, so
. Thus, our product is
, so
. Adding the digits, we have
.
-Sepehr2010, minor edits by the_eaglercraft_grinder
Solution 2
We have 4 integers in our problem. Let's call the smallest of them .
either
or
. So, we have the following:
or
.
The second equation has negative solutions, so we discard it. The first equation has , and so
. If we check
we get
.
is
times
, and
is
times
, so our solution checks out. Multiplying
by
, we get
=>
.
~Arcticturn
Solution 3
From the problems, it follows that
Since both
and
must be integer, we get two equations.
43 & 1 yields (0,0) which is not what we want.
129 & 1 yields (22,21) which is more interesting.
Simplifying the equations, we get:
So, the answer is .
~Technodoggo
Solution 4
Say one factorization is The two cases for the other factorization are
and
We know it must be the first because of AM-GM intuition: lesser factors of a number are closer together than larger factors of a number. (We can also try both and see which works.) Thus,
and we find that
meaning the answer is
~DouDragon
Solution 5
Since we are given that some pairs of divisors differ by 20 and 23 and we can let the pair be and
as well as
and
. We also know the product of both the complementary divisors give the same number so
.
Now we let
. Then we substitute and get
. Finally we multiply by 4 and get
.
Then we use differences of squares and get
+
=129,
-
=1. We finish by getting
65 and
. So
Adding the digits, we have
.
~averageguy
Nunber sense note: To avoid tedious multiplication of 2-digit numbers, observe that , and
, and the sum of the digits of
is
, so the sum of the digits of
is equivalent to
. The only equivalent answer choice is
. ~oinava
Solution 6
can be written
with a positive integer
and
with a positive integer
.
The above equations can be reorganized as
The only solution is and
.
Thus,
.
Therefore,
.
So the sum of the digits of
is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 7
We can write as
or
where
and
are divisors of
Since
we know that
, and we can view this as a quadratic in
Since the solution for must be an integer, the discriminant for this quadratic must be a perfect square and therefore
so
Since the discriminant of this quadratic in must also be a perfect square we know that
which we can simplify as
Since they are both positive integers
and
are factors of
so
and
or
and
These systems of equations give us and
respectively, if we plug our values for
into the equation for
we get
and
respectively. The first equation gives us
or
and the second gives us
or
, since
is positive we know that
and
, therefore the sum of the digits of
is
~SailS
Solution 8 (Trial and Error)
Consider the numbers of the form . Since
is always even,
is even. Thus, for
, we calculate
for even values of
. Then, we check if it can also be represented as a product of numbers that differ by
. Checking, we see that
works. Thus, the answer is
~andliu766
Solution 9
. Factoring,
. Let
because clearly
. Then.
. Note that since
, if
, then the equation is
, so
. Plugging this back, we get
and
. Now we find
as
so the answer is
.
-Magnetoninja
Video Solution 1 by OmegaLearn
Video Solution by epicbird08
~EpicBird08
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.