Difference between revisions of "2018 AMC 10A Problems/Problem 25"
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By geometric series, we have | By geometric series, we have | ||
<cmath>\begin{alignat*}{8} | <cmath>\begin{alignat*}{8} | ||
− | A_n&=a\left(1+10+10^2+\cdots+10^n\right)&&=a\cdot\frac{10^n-1}{9}, \\ | + | 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\left(1+10+10^2+\cdots+10^n\right)&&=b\cdot\frac{10^n-1}{9}, \\ | + | 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\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}. | + | 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 |
Revision as of 13:47, 28 August 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
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:
Note that
is a linear equation with
Since it has at least two solutions of
it has at least two solutions of
We conclude that
must be an identity, so we have 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
If
then
which violates the restrictions. However, if
then
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 4 (If you are running out of time)
Considering this is an AMC 10 and calculators are not allowed, the number of digits in and
,
, probably is not greater than 3. Checking cases with
digits first, we find that if
then there are no solutions with a two digit
. Thus we check the case with
and find that
. Thus if
and
then
for
. If
, then
if
,
, and
. If this is the case then
.
~Dhillonr25
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.