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Difference between revisions of "1982 AHSME Problems/Problem 7"

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== Problem ==
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If the operation <math>x \star y</math> is defined by <math>x \star y = (x+1)(y+1) - 1</math>, then which one of the following is FALSE?
 
If the operation <math>x \star y</math> is defined by <math>x \star y = (x+1)(y+1) - 1</math>, then which one of the following is FALSE?
  
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(A) is true because multiplication is commutative.
 
(A) is true because multiplication is commutative.
  
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(B) is false because we have <math>x\star (y + z) = (x+1)(y+z+1) - 1</math> and <math>x\star y + x\star z = (x+1)(y+1) - 1 + (x+1)(z+1) - 1 = (x+1)(y+z+2) - 2 = (x+1)(y+z+1) - (x+1) - 2 = [(x+1)(y+z+1) - 1] + x</math>, which does not match with the previous expression.
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(B) is false because we have <math>x\star (y + z) = (x+1)(y+z+1) - 1</math> and  
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<math>x\star y + x\star z = (x+1)(y+1) - 1 + (x+1)(z+1) - 1</math>
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<math>(x+1)(y+z+2) - 2</math>
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<math>(x+1)(y+z+1) - (x+1) - 2</math>
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<math>[(x+1)(y+z+1) - 1] + x,</math>
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which does not match with the previous expression.
  
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(C) is true because <math>(x-1)\star (x+1) = ((x-1)+1)((x+1)+1) - 1 = x(x+2) - 1 = x^2 + 2x - 1 = [(x+1)^2 - 1] - 1 = (x\star x) - 1</math> for all <math>x</math>.
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(C) is true because  
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<math>(x-1)\star (x+1) = ((x-1)+1)((x+1)+1) - 1</math>
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<math>x(x+2) - 1</math>
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<math>x^2 + 2x - 1</math>
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<math>[(x+1)^2 - 1] - 1</math>
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<math>(x\star x) - 1</math>
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for all <math>x</math>.
  
 
(D) is true because <math>x\star 0 = (x+1)(0+1) - 1 = (x+1) - 1 = x</math> for all <math>x</math>.
 
(D) is true because <math>x\star 0 = (x+1)(0+1) - 1 = (x+1) - 1 = x</math> for all <math>x</math>.

Latest revision as of 12:28, 16 July 2024

Problem

If the operation $x \star y$ is defined by $x \star y = (x+1)(y+1) - 1$, then which one of the following is FALSE?

$\text{(A)} \ x \star y = y\star x \text{ for all real } x,y. \\ \text{(B)} \ x \star (y + z) = ( x \star y ) + (x \star z) \text{ for all real } x,y, \text{ and } z.\\ \text{(C)} \ (x-1) \star (x+1) = (x \star x) - 1 \text{ for all real } x. \\ \text{(D)} \ x \star 0 = x \text{ for all real } x. \\ \text{(E)} \ x \star (y \star z) = (x \star y) \star z \text{ for all real } x,y, \text{ and } z.$

Solution

(A) is true because multiplication is commutative.

(B) is false because we have $x\star (y + z) = (x+1)(y+z+1) - 1$ and $x\star y + x\star z = (x+1)(y+1) - 1 + (x+1)(z+1) - 1$ $(x+1)(y+z+2) - 2$ $(x+1)(y+z+1) - (x+1) - 2$ $[(x+1)(y+z+1) - 1] + x,$ which does not match with the previous expression.

(C) is true because $(x-1)\star (x+1) = ((x-1)+1)((x+1)+1) - 1$ $x(x+2) - 1$ $x^2 + 2x - 1$ $[(x+1)^2 - 1] - 1$ $(x\star x) - 1$ for all $x$.

(D) is true because $x\star 0 = (x+1)(0+1) - 1 = (x+1) - 1 = x$ for all $x$.

(E) is true because multiplication is associative (and the plus-ones inside the parentheses and the minus-ones outside cancel out).

Therefore, our answer is $\boxed{\textbf{(B)}}$, and we are done.

See also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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