|
|
| Line 1: |
Line 1: |
| − | ==Problem==
| + | #redirect [[FidgetBoss 4000's 2019 Mock AMC 12B Problems/Problem 1]] |
| − | If <math>x</math> and <math>y</math> are positive integers such that their product is <math>64,</math> what is the sum of all distinct values for <math>xy+x+y?</math>
| |
| − | | |
| − | <math>\textbf{(A) }391\qquad\textbf{(B) }392\qquad\textbf{(C) }393\qquad\textbf{(D) }394\qquad\textbf{(E) }395\qquad</math>
| |
| − | | |
| − | ==Solution 1==
| |
| − | Use [[Simon's Favorite Factoring Trick]] to deduce <math>xy+x+y=(x+1)(y+1)-1</math>. We know <math>x</math> and <math>y</math> are positive integers and <math>xy=64</math>. So, we have the following possibilities for the pair <math>(x, y): (1, 64), (2, 32), (4, 16), (8, 8)</math>. Note that we do not have to account for flipping <math>x</math> and <math>y</math> because the question asks for only the <math>\emph{distinct}</math> values for <math>xy+x+y</math>. If <math>(x, y)=(1, 64)</math>, then <math>(x+1)(y+1)-1=2\cdot 65-1=129</math>. If <math>(x, y)=(2, 32)</math>, then <math>3\cdot 33-1=98</math>. If <math>(x, y)=(4, 16)</math>, then <math>5\cdot 17-1=84</math>. If <math>(x, y)=(8, 8)</math>, then <math>9\cdot 9-1=80</math>. The sum of these distinct values is <math>129+98+84+80=227+164=\boxed{391}\textbf{(A)}</math>.
| |
| − | | |
| − | ==See also==
| |
| − | {{AMC12 box|year=2019|ab=B|before=First problem|after=[[FidgetBoss 4000's Mock AMC 12B Problems/Problem 2|Problem 2]]}}
| |
| − | | |
| − | [[Category:Introductory Algebra Problems]]
| |
| − | {{FidgetBoss Notice}}
| |
