Difference between revisions of "2019 AMC 12B Problems/Problem 3"

(Solution 2)
Line 1: Line 1:
 
==Problem==
 
==Problem==
If (n+1)!+(n+2)! = n!*440, what is the sum of the digits of n?
+
Which of the following rigid transformations (isometries) maps the line segment <math>\overline{AB}</math> onto the line segment <math>\overline{A'B'}</math> so that the image of <math>A(-2, 1)</math> is <math>A'(2, -1)</math> and the image of <math>B(-1, 4)</math> is <math>B'(1, -4)</math>?
  
==Solution==
+
(A) reflection in the <math>y</math>-axis
n=19
 
sum is 10 (SuperWill)
 
  
 +
(B) counterclockwise rotation around the origin by <math>90^\circ</math>
  
==Solution 2==
+
(C) translation by <math>3</math> units to the right and <math>5</math> units down
  
Divide both sides by <math>n!</math>:
+
(D) reflection in the <math>x</math>-axis
  
 +
(E) clockwise rotation around the origin by <math>180^\circ</math>
  
<math>(n+1)+(n+1)(n+2)=440</math>
+
==Solution==
 
 
factor out <math>(n+1)</math>:
 
 
 
<math>(n+1)*(n+3)=440</math>
 
 
 
  
prime factorization of <math>440</math> and a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>\boxed{n=19}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{AMC12 box|year=2019|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:28, 14 February 2019

Problem

Which of the following rigid transformations (isometries) maps the line segment $\overline{AB}$ onto the line segment $\overline{A'B'}$ so that the image of $A(-2, 1)$ is $A'(2, -1)$ and the image of $B(-1, 4)$ is $B'(1, -4)$?

(A) reflection in the $y$-axis

(B) counterclockwise rotation around the origin by $90^\circ$

(C) translation by $3$ units to the right and $5$ units down

(D) reflection in the $x$-axis

(E) clockwise rotation around the origin by $180^\circ$

Solution

See Also

2019 AMC 12B (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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png