Difference between revisions of "2019 AMC 10A Problems/Problem 5"

Revision as of 16:43, 9 February 2019

The following problem is from both the 2019 AMC 10A #5 and 2019 AMC 12A #4, so both problems redirect to this page.

What is the greatest number of consecutive integers whose sum is $45?$

$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$

Note that every term in the sequence $-44, -43..., 44, 45$ cancels out except $45$ . This results in $\boxed{\textbf{(D) } 90 }$ integers.

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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

@@ Line 1: / Line 1: @@
-==Problem 5==
+{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #5]] and [[2019 AMC 12A Problems|2019 AMC 12A #4]]}}
-What is the greatest number of consecutive integers whose sum is <math>45 ?</math>
+==Problem==
+What is the greatest number of consecutive integers whose sum is <math>45?</math>
 <math>\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120</math>
@@ Line 6: / Line 8: @@
 ==Solution==
 Note that every term in the sequence <math>-44, -43..., 44, 45</math> cancels out except <math>45</math>. This results in <math>\boxed{\textbf{(D) } 90 }</math> integers.
+==See Also==
+{{AMC10 box|year=2019|ab=A|num-b=4|num-a=6}}
+{{AMC12 box|year=2019|ab=A|num-b=3|num-a=5}}
+{{MAA Notice}}