2024 AMC 10A Problems/Problem 15
- The following problem is from both the 2024 AMC 10A #15 and 2024 AMC 12A #9, so both problems redirect to this page.
Contents
[hide]Problem
Let be the greatest integer such that both and are perfect squares. What is the units digit of ?
Solution 1
Let and for some positive integers and We subtract the first equation from the second, then apply the difference of squares: Note that and have the same parity, and
We wish to maximize both and so we maximize and minimize It follows that from which
Finally, we get so the units digit of is
~MRENTHUSIASM ~Tacos_are_yummy_1
Solution 2 (not rigorously proven)
We assume that when the maximum values are achieved, the two squares are consecutive squares. However, since both have the same parity, the closest we can get to this is that they are 1 square apart.
Let the square between and be . So, we have and . Subtracting the two, we have , which yields , which leads to . Therefore, the two squares are and , which both have units digit . Since both and have units digit , will have units digit .
~i_am_suk_at_math_2
Solution 2a (perhaps more clear)
Ideally, and are close squares. We would want them to be consecutive squares (squares of consecutive numbers), but and are of the same parity (and so are their squares) while consecutive squares are always opposite parity, so this is impossible. The next-best option is and ; the latter expands into , and subtracting the former from this result, we see that . Then .
~Technodoggo
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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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