Difference between revisions of "1982 AHSME Problems/Problem 7"
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(A) is true because multiplication is commutative. | (A) is true because multiplication is commutative. | ||
− | (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 | + | (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</math> | ||
+ | <math>(x+1)(y+z+2) - 2</math> | ||
+ | <math>(x+1)(y+z+1) - (x+1) - 2</math> | ||
+ | <math>[(x+1)(y+z+1) - 1] + x,</math> | ||
+ | which does not match with the previous expression. | ||
− | (C) is true because <math>(x-1)\star (x+1) = ((x-1)+1)((x+1)+1) - 1 | + | (C) is true because |
+ | <math>(x-1)\star (x+1) = ((x-1)+1)((x+1)+1) - 1</math> | ||
+ | <math>x(x+2) - 1</math> | ||
+ | <math>x^2 + 2x - 1</math> | ||
+ | <math>[(x+1)^2 - 1] - 1</math> | ||
+ | <math>(x\star x) - 1</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>. | (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 22:00, 24 September 2020
If the operation is defined by , then which one of the following is FALSE?
Solution
(A) is true because multiplication is commutative.
(B) is false because we have and which does not match with the previous expression.
(C) is true because for all .
(D) is true because for all .
(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 , 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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