Difference between revisions of "1982 AHSME Problems/Problem 11"
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| − | == Problem | + | == Problem == |
| + | |||
| + | How many integers with four different digits are there between <math>1,000</math> and <math>9,999</math> such that the absolute value of | ||
| + | the difference between the first digit and the last digit is <math>2</math>? | ||
| + | |||
| + | <math>\textbf {(A)}\ 672 \qquad | ||
| + | \textbf {(B)}\ 784 \qquad | ||
| + | \textbf {(C)}\ 840 \qquad | ||
| + | \textbf {(D)}\ 896 \qquad | ||
| + | \textbf {(E)}\ 1008</math> | ||
| + | |||
| + | == 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>15\cdot56 = \boxed {\left(C\right) 840}</math> numbers that fulfill these circumstances. | 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. | ||
Latest revision as of 13:32, 16 July 2024
Problem
How many integers with four different digits are there between 
and 
such that the absolute value of
the difference between the first digit and the last digit is 
?
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.
