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

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== Problem 11 Solution ==
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== Problem ==
  
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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.
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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
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the difference between the first digit and the last digit is <math>2</math>?
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<math>\textbf {(A)}\ 672 \qquad
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\textbf {(B)}\ 784 \qquad
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\textbf {(C)}\ 840 \qquad
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\textbf {(D)}\ 896 \qquad
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\textbf {(E)}\ 1008</math>   
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== Solution ==
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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>15\cdot56 = \boxed {\left(C\right) 840}</math> numbers that fulfill these circumstances.

Latest revision as of 12:32, 16 July 2024

Problem

How many integers with four different digits are there between $1,000$ and $9,999$ such that the absolute value of the difference between the first digit and the last digit is $2$?

$\textbf {(A)}\ 672 \qquad \textbf {(B)}\ 784 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 896 \qquad \textbf {(E)}\ 1008$

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 $15\cdot56 = \boxed {\left(C\right) 840}$ numbers that fulfill these circumstances.

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