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 |
Revision as of 19:48, 21 October 2021
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