Difference between revisions of "2018 AMC 10A Problems/Problem 25"
MRENTHUSIASM (talk | contribs) (Made the argument in Sol 1 more rigorous. Previously Sol 1 says that plugging in n = 0, but n = 0 is not a valid value. Also made some cosmetic changes to Sol 1 and made more intermediate steps.) |
MRENTHUSIASM (talk | contribs) m (→Solution 3) |
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− | {{duplicate|[[2018 AMC | + | {{duplicate|[[2018 AMC 10A Problems/Problem 25|2018 AMC 10A #25]] and [[2018 AMC 12A Problems/Problem 25|2018 AMC 12A #25]]}} |
== Problem == | == Problem == | ||
Line 10: | Line 10: | ||
By geometric series, we have | By geometric series, we have | ||
<cmath>\begin{alignat*}{8} | <cmath>\begin{alignat*}{8} | ||
− | A_n&=a\ | + | A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\ |
− | B_n&=b\ | + | B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\right)&&=b\cdot\frac{10^n-1}{9}, \\ |
− | C_n&=c\ | + | C_n&=c\bigl(\phantom{ }\underbrace{111\cdots1}_{2n\text{ digits}}\phantom{ }\bigr)&&=c\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}. |
\end{alignat*}</cmath> | \end{alignat*}</cmath> | ||
By substitution, we rewrite the given equation <math>C_n - B_n = A_n^2</math> as | By substitution, we rewrite the given equation <math>C_n - B_n = A_n^2</math> as | ||
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\left(9c-a^2\right)10^n &= 9b-9c-a^2. &&(\bigstar) | \left(9c-a^2\right)10^n &= 9b-9c-a^2. &&(\bigstar) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Note that <math>(\bigstar)</math> is a linear equation with <math> | + | Let <math>y=10^n.</math> Note that <math>(\bigstar)</math> is a linear equation with <math>y,</math> and <math>y</math> is a one-to-one function of <math>n.</math> Since <math>(\bigstar)</math> has at least two solutions of <math>n,</math> it has at least two solutions of <math>y.</math> We conclude that <math>(\bigstar)</math> must be an identity, so we get the following system of equations: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
9c-a^2&=0, \\ | 9c-a^2&=0, \\ | ||
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The first equation implies that <math>c=\frac{a^2}{9}.</math> Substituting this into the second equation gives <math>b=\frac{2a^2}{9}.</math> | The first equation implies that <math>c=\frac{a^2}{9}.</math> Substituting this into the second equation gives <math>b=\frac{2a^2}{9}.</math> | ||
− | To maximize <math>a + b + c = a + \frac{a^2}{3},</math> we need to maximize <math>a.</math> Clearly, <math>a</math> must be divisible by <math>3.</math> | + | To maximize <math>a + b + c = a + \frac{a^2}{3},</math> we need to maximize <math>a.</math> Clearly, <math>a</math> must be divisible by <math>3.</math> The possibilities for <math>(a,b,c)</math> are <math>(9,18,9),(6,8,4),</math> or <math>(3,2,1),</math> but <math>(9,18,9)</math> is invalid. Therefore, the greatest possible value of <math>a + b + c</math> is <math>6+8+4=\boxed{\textbf{(D) } 18}.</math> |
+ | |||
+ | ~CantonMathGuy (Solution) | ||
+ | |||
+ | ~MRENTHUSIASM (Revision) | ||
== Solution 2== | == Solution 2== | ||
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We now only need to know whether <math>a + b + c = 20</math> might work for any larger <math>n</math>. We will always get equations like <math>100001c - b = 11111a^2</math> where the <math>c</math> coefficient is very close to being nine times the <math>a</math> coefficient. Since the <math>b</math> term will be quite insignificant, we know that once again <math>a^2</math> must equal <math>9c</math>, and thus <math>a = 9, c = 9</math> is our only hope to reach <math>20</math>. Substituting and dividing through by <math>9</math>, we will have something like <math>100001 - \frac{b}{9} = 99999</math>. No matter what <math>n</math> really was, <math>b</math> is out of range (and certainly isn't <math>2</math> as we would have needed). | We now only need to know whether <math>a + b + c = 20</math> might work for any larger <math>n</math>. We will always get equations like <math>100001c - b = 11111a^2</math> where the <math>c</math> coefficient is very close to being nine times the <math>a</math> coefficient. Since the <math>b</math> term will be quite insignificant, we know that once again <math>a^2</math> must equal <math>9c</math>, and thus <math>a = 9, c = 9</math> is our only hope to reach <math>20</math>. Substituting and dividing through by <math>9</math>, we will have something like <math>100001 - \frac{b}{9} = 99999</math>. No matter what <math>n</math> really was, <math>b</math> is out of range (and certainly isn't <math>2</math> as we would have needed). | ||
− | The answer then is <math>\boxed{\textbf{(D)} | + | The answer then is <math>\boxed{\textbf{(D) } 18}</math>. |
== Solution 3 == | == Solution 3 == | ||
− | The given equation can be written as | + | The given equation can be written as |
− | <cmath>c \cdot ( \overbrace{1111 \ldots 1111}^\text{ | + | <cmath>c \cdot (\phantom{ } \overbrace{1111 \ldots 1111}^{2n\text{ digits}}\phantom{ }) - b \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ })^2.</cmath> |
− | Divide by <math>\overbrace{11 \ldots 11}^\text{ | + | Divide by <math>\overbrace{11 \ldots 11}^{n\text{ digits}}</math> on both sides: |
− | <cmath>c \cdot ( \overbrace{1000 \ldots 0001}^ | + | <cmath>c \cdot (\phantom{ } \overbrace{1000 \ldots 0001}^{n+1\text{ digits}}\phantom{ }) - b = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }).</cmath> |
− | Next, split the first term to make it easier to deal with | + | Next, split the first term to make it easier to deal with: |
− | <cmath>2c + c \cdot (\overbrace{99 \ldots 99}^\text{ | + | <cmath>\begin{align*} |
− | + | 2c + c \cdot (\phantom{ }\overbrace{99 \ldots 99}^{n\text{ digits}}\phantom{ }) - b &= a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) \\ | |
− | Because <math>2c - b</math> and <math>a^2 - 9c</math> are constants and because there must be at least two distinct values of <math>n</math> that satisfy, <math>2c - b = a^2 - 9c = 0</math> | + | 2c - b &= (a^2 - 9c) \cdot (\phantom{ }\overbrace{11 \ldots 11}^{n\text{ digits}}\phantom{ }). |
− | <cmath>2c=b | + | \end{align*}</cmath> |
− | + | Because <math>2c - b</math> and <math>a^2 - 9c</math> are constants and because there must be at least two distinct values of <math>n</math> that satisfy, <math>2c - b = a^2 - 9c = 0.</math> Thus, we have | |
− | Knowing that <math>a | + | <cmath>\begin{align*} |
+ | 2c&=b, \\ | ||
+ | a^2&=9c. | ||
+ | \end{align*}</cmath> | ||
+ | Knowing that <math>a,b,</math> and <math>c</math> are single digit positive integers and that <math>9c</math> must be a perfect square, the values of <math>(a,b,c)</math> that satisfy both equations are <math>(3,2,1)</math> and <math>(6,8,4).</math> Finally, <math>6 + 8 + 4 = \boxed{\textbf{(D) } 18}.</math> | ||
− | ~LegionOfAvatars | + | ~LegionOfAvatars (Solution) |
− | + | ~MRENTHUSIASM (Reformatting) | |
− | + | ==Solution 4== | |
− | + | By [[PaperMath’s sum]], the answer is <math>6+8+4=\boxed{\textbf{(D) } 18}</math> | |
== Video Solution by Richard Rusczyk == | == Video Solution by Richard Rusczyk == |
Latest revision as of 16:50, 21 January 2024
- The following problem is from both the 2018 AMC 10A #25 and 2018 AMC 12A #25, so both problems redirect to this page.
Contents
Problem
For a positive integer and nonzero digits
,
, and
, let
be the
-digit integer each of whose digits is equal to
; let
be the
-digit integer each of whose digits is equal to
, and let
be the
-digit (not
-digit) integer each of whose digits is equal to
. What is the greatest possible value of
for which there are at least two values of
such that
?
Solution 1
By geometric series, we have
By substitution, we rewrite the given equation
as
Since
it follows that
We divide both sides by
and then rearrange:
Let
Note that
is a linear equation with
and
is a one-to-one function of
Since
has at least two solutions of
it has at least two solutions of
We conclude that
must be an identity, so we get the following system of equations:
The first equation implies that
Substituting this into the second equation gives
To maximize we need to maximize
Clearly,
must be divisible by
The possibilities for
are
or
but
is invalid. Therefore, the greatest possible value of
is
~CantonMathGuy (Solution)
~MRENTHUSIASM (Revision)
Solution 2
Immediately start trying and
. These give the system of equations
and
(which simplifies to
). These imply that
, so the possible
pairs are
,
, and
. The first puts
out of range but the second makes
. We now know the answer is at least
.
We now only need to know whether might work for any larger
. We will always get equations like
where the
coefficient is very close to being nine times the
coefficient. Since the
term will be quite insignificant, we know that once again
must equal
, and thus
is our only hope to reach
. Substituting and dividing through by
, we will have something like
. No matter what
really was,
is out of range (and certainly isn't
as we would have needed).
The answer then is .
Solution 3
The given equation can be written as
Divide by
on both sides:
Next, split the first term to make it easier to deal with:
Because
and
are constants and because there must be at least two distinct values of
that satisfy,
Thus, we have
Knowing that
and
are single digit positive integers and that
must be a perfect square, the values of
that satisfy both equations are
and
Finally,
~LegionOfAvatars (Solution)
~MRENTHUSIASM (Reformatting)
Solution 4
By PaperMath’s sum, the answer is
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2018amc10a/470
~ dolphin7
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.