Difference between revisions of "1982 AHSME Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | We simply ignore the floor of <math>x</math>. Then, we have <math>y</math> = <math>2x + 3</math> = <math>3(x-2)+5</math>. Solving for <math>3x - 1 = 2x + 3</math>, we get <math>x = 4</math>. For the floor of <math>x</math>, we have <math>x</math> is between <math>4</math> and <math>5</math>. Plugging in <math>8</math> + <math>3</math> = <math>11</math> for <math>y</math>, we have <math>y = 11</math>. We have <math>11 + 4.x</math> = | + | We simply ignore the floor of <math>x</math>. Then, we have <math>y</math> = <math>2x + 3</math> = <math>3(x-2)+5</math>. Solving for <math>3x - 1 = 2x + 3</math>, we get <math>x = 4</math>. For the floor of <math>x</math>, we have <math>x</math> is between <math>4</math> and <math>5</math>. Plugging in <math>8</math> + <math>3</math> = <math>11</math> for <math>y</math>, we have <math>y = 11</math>. We have <math>11 + 4.x</math> = <math>\boxed {(D)}</math> |
~Arcticturn | ~Arcticturn | ||
+ | |||
+ | == Solution 2 (RIGID) == | ||
+ | Since <math>x</math> is not an integer, we let <math>x=a+b</math>, where <math>0<b<1</math>. | ||
+ | |||
+ | So <math>2[x]+3=2a+3</math>. <math>3[x-2]+5=3a-1</math>. | ||
+ | |||
+ | <math>2a+3=3a-1</math>. <math>a=4</math>. So we know that <math>x</math> is between 4 and 5. <math>y=11</math>. So <math>x+y</math> is between <math>15</math> and <math>16</math>. Select <math>\boxed{D}</math>. | ||
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+ | ~Alvie567 |
Latest revision as of 21:41, 16 August 2022
Problem
Let denote the greatest integer not exceeding . Let and satisfy the simultaneous equations
If is not an integer, then is
Solution
We simply ignore the floor of . Then, we have = = . Solving for , we get . For the floor of , we have is between and . Plugging in + = for , we have . We have =
~Arcticturn
Solution 2 (RIGID)
Since is not an integer, we let , where .
So . .
. . So we know that is between 4 and 5. . So is between and . Select .
~Alvie567