Difference between revisions of "2018 AMC 10A Problems/Problem 25"
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The answer then is <math>\boxed{\textbf{(D)} \text{ 18}}</math>. | The answer then is <math>\boxed{\textbf{(D)} \text{ 18}}</math>. | ||
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+ | == Solution 3 == | ||
+ | Note that <math>(0.\overline{6})^2 = 0.\overline{4}</math>. Setting <math>a = 6</math> and <math>c = 4</math>, we see that the solution <math>(a, b, c) = (6, 8, 4)</math> works for all possible values of <math>n</math>. The answer is <math>6 + 8 + 4 = {\boxed{\textbf{(D)} \text{ 18}}}</math>. | ||
==See Also== | ==See Also== |
Revision as of 13:32, 18 February 2018
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. Matching coefficients, we need
To maximize
, we need to maximize
. Since
and
must be integers,
must be a multiple of 3. If
then
exceeds 9. However, if
then
and
for an answer of
. (CantonMathGuy)
Solution 2 (quicker?)
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
Note that . Setting
and
, we see that the solution
works for all possible values of
. The answer is
.
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 |
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