Difference between revisions of "1982 AHSME Problems/Problem 11"
(→Problem 11 Solution) |
(→Problem 11 Solution) |
||
| Line 1: | Line 1: | ||
== Problem 11 Solution == | == 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 <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> | + | 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>15\cdot56 = \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 
and 
or 
and 
Substituting we have 
and 
Thus 
and 
which yields 16 pairs (x, y) such that the absolute value of the difference between the x and y is 
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 
ways to fill the middle 2 places with digits, thus by the multiplication principles (counting) there are 
numbers that fulfill these circumstances.
