Art of Problem Solving

2024 AMC 10A Problems/Problem 15: Difference between revisions

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==Problem==
Let <math>M</math> be the greatest integer such that both <math>M+1213</math> and <math>M+3773</math> are perfect squares. What is the units digit of <math>M</math>?


<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }8</math>
==Solution==
==See also==
{{AMC10 box|year=2024|ab=A|num-b=2|num-a=4}}
{{MAA Notice}}

Revision as of 15:58, 8 November 2024

Problem

Let $M$ be the greatest integer such that both $M+1213$ and $M+3773$ are perfect squares. What is the units digit of $M$?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }8$

Solution

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

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