Difference between revisions of "2023 AMC 10B Problems/Problem 14"
Grolarbear (talk | contribs) (→Solution 4 (Nice Substitution)) |
Icyfire500 (talk | contribs) m (→Solution 1) |
||
Line 17: | Line 17: | ||
This basically say that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square which is practically impossible except <math>mn=0</math> or <math>mn+1=0</math>. | This basically say that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square which is practically impossible except <math>mn=0</math> or <math>mn+1=0</math>. | ||
<math>mn=0</math> gives <math>(0,0)</math>. | <math>mn=0</math> gives <math>(0,0)</math>. | ||
− | <math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>. | + | <math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>. Answer: <math>\boxed{(C)3}.</math> |
~Technodoggo ~minor edits by lucaswujc | ~Technodoggo ~minor edits by lucaswujc |
Revision as of 15:15, 29 November 2023
Contents
[hide]Problem
How many ordered pairs of integers satisfy the equation
?
Solution 1
Clearly, is 1 solution. However there are definitely more, so we apply Simon's Favorite Factoring Expression to get this:
This basically say that the product of two consecutive numbers must be a perfect square which is practically impossible except
or
.
gives
.
gives
. Answer:
~Technodoggo ~minor edits by lucaswujc
Solution 2
Case 1: .
In this case, .
Case 2: .
Denote .
Denote
and
.
Thus,
.
Thus, the equation given in this problem can be written as
Modulo , we have
.
Because
, we must have
.
Plugging this into the above equation, we get
.
Thus, we must have
and
.
Thus, there are two solutions in this case: and
.
Putting all cases together, the total number of solutions is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Discriminant)
We can move all terms to one side and wrote the equation as a quadratic in terms of to get
The discriminant of this quadratic is
For
to be an integer, we must have
be a perfect square. Thus, either
is a perfect square or
and
. The first case gives
, which result in the equations
and
, for a total of two pairs:
and
. The second case gives the equation
, so it's only pair is
. In total, the total number of solutions is
.
~A_MatheMagician
Solution 4 (Nice Substitution)
Let then
Completing the square then gives
Since the RHS is a square, clearly the only solutions are
and
. The first gives
while the second gives
and
by solving it as a quadratic with roots
and
. Thus there are
solutions.
~ Grolarbear
Video Solution by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.