Difference between revisions of "2018 AMC 10A Problems/Problem 25"
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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}, \\ | 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}, \\ |
Revision as of 11:27, 29 October 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: 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)
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.