Difference between revisions of "1985 AHSME Problems/Problem 30"
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<cmath> 4a^2-32a+55>0 </cmath> | <cmath> 4a^2-32a+55>0 </cmath> | ||
− | The first inequality simplifies to <math>(2a-10)^2\le 49</math>, or <math>|2a-10|\le 7</math>. Since <math>2a-10</math> is even, we must have <math>|2a-10|\le 6</math>, so <math>|a-5|\le 3</math>. Therefore, <math>2\le a\le 8</math>. The second inequality simplifies to <math>(2a-8)^2 > 9</math>, or <math>|2a-8| > 3</math>. Therefore, as <math>2a-8</math> is even, we have <math>|2a-8|\ge 4</math>, or <math>|a-4|\ge 2</math>. Hence <math>a\ge 6</math> or <math>a\le 2</math> Since both inequalities must be satisfied, we see that only <math>a=2</math>, <math>a=6</math>, <math>a=7</math>, and <math>a=8</math> satisfy both inequalities. Each will lead to a distinct solution for <math>n</math>, and thus for <math>x</math>, for a total of <math>4</math> positive solutions. | + | The first inequality simplifies to <math>(2a-10)^2\le 49</math>, or <math>|2a-10|\le 7</math>. Since <math>2a-10</math> is even, we must have <math>|2a-10|\le 6</math>, so <math>|a-5|\le 3</math>. Therefore, <math>2\le a\le 8</math>. The second inequality simplifies to <math>(2a-8)^2 > 9</math>, or <math>|2a-8| > 3</math>. Therefore, as <math>2a-8</math> is even, we have <math>|2a-8|\ge 4</math>, or <math>|a-4|\ge 2</math>. Hence <math>a\ge 6</math> or <math>a\le 2</math>. Since both inequalities must be satisfied, we see that only <math>a=2</math>, <math>a=6</math>, <math>a=7</math>, and <math>a=8</math> satisfy both inequalities. Each will lead to a distinct solution for <math>n</math>, and thus for <math>x</math>, for a total of <math>4</math> positive solutions. |
Now let <math> x=-\frac{\sqrt{n}}{2} </math>. We have | Now let <math> x=-\frac{\sqrt{n}}{2} </math>. We have |
Revision as of 00:22, 3 April 2018
Problem
Let be the greatest integer less than or equal to
. Then the number of real solutions to
is
Solution
We can rearrange the equation into . Obviously, the RHS is an integer, so
for some integer
. We can therefore make the substitution
to get
(We'll try the case where later.) Now let
for a nonnegative integer
, so that
.
Going back to , this implies
. Making the substitution
gives the system of inequalities
The first inequality simplifies to , or
. Since
is even, we must have
, so
. Therefore,
. The second inequality simplifies to
, or
. Therefore, as
is even, we have
, or
. Hence
or
. Since both inequalities must be satisfied, we see that only
,
,
, and
satisfy both inequalities. Each will lead to a distinct solution for
, and thus for
, for a total of
positive solutions.
Now let . We have
Since , this can be rewritten as
Since is positive, the least possible value of
is
, hence
, i.e., it must be negative. But it is also equal to
, which is positive, and this is a contradiction. Therefore, there are no negative roots.
The total number of roots to this equation is thus , or
.
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Problem | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.