Difference between revisions of "1998 AHSME Problems/Problem 2"

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<math> \mathrm{(A) \  }13 \qquad \mathrm{(B) \  }14 \qquad \mathrm{(C) \  } 15\qquad \mathrm{(D) \  }16 \qquad \mathrm{(E) \  } 17 </math>
 
<math> \mathrm{(A) \  }13 \qquad \mathrm{(B) \  }14 \qquad \mathrm{(C) \  } 15\qquad \mathrm{(D) \  }16 \qquad \mathrm{(E) \  } 17 </math>
  
[[1998 AHSME Problems/Problem 2|Solution]]
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==Solution==
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If we want <math>\frac{A+B}{C+D}</math> to be as large as possible, we want to try to maximize the numerator <math>A+B</math> and minimize the denominator <math>C+D</math>.  Picking <math>A=9</math> and <math>B=8</math> will maximize the numerator, and picking <math>C=0</math> and <math>D=1</math> will minimize the denominator. 
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Checking to make sure the fraction is an integer, <math>\frac{A+B}{C+D} = \frac{17}{1} = 17</math>, and so the values are correct, and <math>A+B = 17</math>, giving the answer <math>\boxed{E}</math>.
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== See also ==
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{{AHSME box|year=1998|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 14:28, 5 July 2013

Problem 2

Letters $A,B,C,$ and $D$ represent four different digits selected from $0,1,2,\ldots ,9.$ If $(A+B)/(C+D)$ is an integer that is as large as possible, what is the value of $A+B$?

$\mathrm{(A) \  }13 \qquad \mathrm{(B) \  }14 \qquad \mathrm{(C) \  } 15\qquad \mathrm{(D) \  }16 \qquad \mathrm{(E) \  } 17$

Solution

If we want $\frac{A+B}{C+D}$ to be as large as possible, we want to try to maximize the numerator $A+B$ and minimize the denominator $C+D$. Picking $A=9$ and $B=8$ will maximize the numerator, and picking $C=0$ and $D=1$ will minimize the denominator.

Checking to make sure the fraction is an integer, $\frac{A+B}{C+D} = \frac{17}{1} = 17$, and so the values are correct, and $A+B = 17$, giving the answer $\boxed{E}$.

See also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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