Difference between revisions of "2018 AMC 10A Problems/Problem 25"
m (→Solution 1) |
(Add solution 4) |
||
Line 23: | Line 23: | ||
== Solution 3 (Cheating) == | == Solution 3 (Cheating) == | ||
+ | |||
Notice that <math>(0.\overline{3})^2 = 0.\overline{1}</math> and <math>(0.\overline{6})^2 = 0.\overline{4}</math>. Setting <math>a = 3</math> and <math>c = 1</math>, we see <math>b = 2</math> works for all possible values of <math>n</math>. Similarly, if <math>a = 6</math> and <math>c = 4</math>, then <math>b = 8</math> works for all possible values of <math>n</math>. The second solution yields a greater sum of <math>\boxed{\textbf{(D)} \text{ 18}}</math>. | Notice that <math>(0.\overline{3})^2 = 0.\overline{1}</math> and <math>(0.\overline{6})^2 = 0.\overline{4}</math>. Setting <math>a = 3</math> and <math>c = 1</math>, we see <math>b = 2</math> works for all possible values of <math>n</math>. Similarly, if <math>a = 6</math> and <math>c = 4</math>, then <math>b = 8</math> works for all possible values of <math>n</math>. The second solution yields a greater sum of <math>\boxed{\textbf{(D)} \text{ 18}}</math>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | The given equation can be written as: | ||
+ | <cmath>c \cdot ( \overbrace{1111 \ldots 1111}^\text{2n}) - b \cdot ( \overbrace{11 \ldots 11}^\text{n} ) = a^2 \cdot ( \overbrace{11 \ldots 11}^\text{n} )^2</cmath> | ||
+ | Divide by <math>\overbrace{11 \ldots 11}^\text{n}</math> on both sides: | ||
+ | <cmath>c \cdot ( \overbrace{1000 \ldots 0001}^\text{n+1}) - b = a^2 \cdot ( \overbrace{11 \ldots 11}^\text{n} )</cmath> | ||
+ | Next, split the first term to make it easier to deal with. | ||
+ | <cmath>2c + c \cdot (\overbrace{99 \ldots 99}^\text{n}) - b = a^2 \cdot ( \overbrace{11 \ldots 11}^\text{n} )</cmath> | ||
+ | <cmath>2c - b = (a^2 - 9c) \cdot (\overbrace{11 \ldots 11}^\text{n})</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: | ||
+ | <cmath>2c=b</cmath> | ||
+ | <cmath>a^2=9c</cmath> | ||
+ | Knowing that <math>a</math>, <math>b</math>, and <math>c</math> are single digit positive integers and that <math>9c</math> must be a perfect square, the possible 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)} \text{18}}</math>. | ||
+ | |||
+ | ~LegionOfAvatars | ||
== Video Solution by Richard Rusczyk == | == Video Solution by Richard Rusczyk == |
Revision as of 22:55, 11 January 2021
- The following problem is from both the 2018 AMC 12A #25 and 2018 AMC 10A #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
Observe ; similarly,
and
. The relation
rewrites as
Since
,
and we may cancel out a factor of
to obtain
This is a linear equation in
. Thus, if two distinct values of
satisfy it, then all values of
will. Now we plug in
and
(or some other number), we get
and
. Solving the equations for
and
, we get
To maximize
, we need to maximize
. Since
and
must be integers,
must be a multiple of
. If
then
exceeds
. However, if
then
and
for an answer of
.
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 (Cheating)
Notice that and
. Setting
and
, we see
works for all possible values of
. Similarly, if
and
, then
works for all possible values of
. The second solution yields a greater sum of
.
Solution 4
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 possible values of
that satisfy both equations are
and
Finally,
.
~LegionOfAvatars
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.