Difference between revisions of "1982 AHSME Problems/Problem 15"
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+ | == 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 22:41, 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