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Difference between revisions of "1982 AHSME Problems/Problem 11"

(→‎Problem 11 Solution)
(→‎Problem 11 Solution)
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== Problem 11 Solution ==
 
== Problem 11 Solution ==
  
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Since <math>BO</math> and <math>CO</math> are angle bisectors of angles <math>B</math> and <math>C</math> respectively, <math>\angle MBO = \angle OBC</math> and similarly <math>\angle NCO = \angle OCB</math>. Because <math>MN</math> and <math>BC</math> are parallel, <math>\angle OBC = \angle MOB</math> and <math>\angle NOC = \angle OCB</math> by corresponding angles. This relation makes <math>\triangle MOB</math> and <math>\triangle NOC</math> isosceles. This makes <math>MB = MO</math> and <math>NO = NC</math>. Therefore the perimeter of <math>\triangle AMN</math> is <math>12 + 18 = 30</math>.
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All of the digits of the numbers must be 0 to 9. If the first and last digits are x and y, we have <math>x-y=2</math> and <math>0<x<9</math> or <math>y-x=2,</math> and <math>0<y<9.</math> Substituting we have <math>0<x+2<9,</math> and <math>0<y+2<9.</math> Thus <math>0<x<7</math> and <math>0<y<7,</math> which yields 16 pairs (x, y) such that the absolute value of the difference between the x and y is <math>2.</math> However, we are not done. If 0 is the last digit ( with the pair (0, 2):) then we won't have a 4 digit number, so our real value is 15. Because our digits are distinct, there are <math>(10-2)(10-3)</math> ways to fill the middle 2 places with digits, thus by the multiplication principles (counting) there are <math>15x56 = \boxed {\left(C\right) 840}</math> numbers that fulfill these circumstances.

Revision as of 00:14, 12 October 2020

Problem 11 Solution

All of the digits of the numbers must be 0 to 9. If the first and last digits are x and y, we have $x-y=2$ and $0<x<9$ or $y-x=2,$ and $0<y<9.$ Substituting we have $0<x+2<9,$ and $0<y+2<9.$ Thus $0<x<7$ and $0<y<7,$ which yields 16 pairs (x, y) such that the absolute value of the difference between the x and y is $2.$ However, we are not done. If 0 is the last digit ( with the pair (0, 2):) then we won't have a 4 digit number, so our real value is 15. Because our digits are distinct, there are $(10-2)(10-3)$ ways to fill the middle 2 places with digits, thus by the multiplication principles (counting) there are $15x56 = \boxed {\left(C\right) 840}$ numbers that fulfill these circumstances.

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