Difference between revisions of "2010 AIME I Problems/Problem 5"
m |
Jerry122805 (talk | contribs) (→Solution 1) |
||
(9 intermediate revisions by 4 users not shown) | |||
Line 3: | Line 3: | ||
Positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> satisfy <math>a > b > c > d</math>, <math>a + b + c + d = 2010</math>, and <math>a^2 - b^2 + c^2 - d^2 = 2010</math>. Find the number of possible values of <math>a</math>. | Positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> satisfy <math>a > b > c > d</math>, <math>a + b + c + d = 2010</math>, and <math>a^2 - b^2 + c^2 - d^2 = 2010</math>. Find the number of possible values of <math>a</math>. | ||
− | ==Solution== | + | == Solution 1 == |
+ | Using the [[difference of squares]], <math>2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010</math>, where equality must hold so <math>b = a - 1</math> and <math>d = c - 1</math>. Then we see <math>a = 1004</math> is maximal and <math>a = 504</math> is minimal, so the answer is <math>\boxed{501}</math>. | ||
+ | |||
+ | |||
+ | Note: We can also find that <math>b=a-1</math> in another way. We know <cmath>a^{2} - b^{2} + c^{2} - d^{2}=(a+b)+(c+d) \implies (a+b)(a-b)-(a+b)+(c+d)(c-d)-(c+d)=0 \implies (a+b)(a-b-1)+(c+d)(c-d-1)=0</cmath> | ||
− | + | Therefore, one of <math>(a+b)(a-b-1)</math> and <math>(c+d)(c-d-1)</math> must be <math>0.</math> Clearly, <math>a+b \neq 0</math> since then one would be positive and negative, or both would be zero. Therefore, <math>a-b-1=0</math> so <math>a=b+1</math>. Similarly, we can deduce that <math>c=d+1.</math> | |
− | |||
− | + | == Solution 2 == | |
Since <math>a+b</math> must be greater than <math>1005</math>, it follows that the only possible value for <math>a-b</math> is <math>1</math> (otherwise the quantity <math>a^2 - b^2</math> would be greater than <math>2010</math>). Therefore the only possible ordered pairs for <math>(a,b)</math> are <math>(504, 503)</math>, <math>(505, 504)</math>, ... , <math>(1004, 1003)</math>, so <math>a</math> has <math>\boxed{501}</math> possible values. | Since <math>a+b</math> must be greater than <math>1005</math>, it follows that the only possible value for <math>a-b</math> is <math>1</math> (otherwise the quantity <math>a^2 - b^2</math> would be greater than <math>2010</math>). Therefore the only possible ordered pairs for <math>(a,b)</math> are <math>(504, 503)</math>, <math>(505, 504)</math>, ... , <math>(1004, 1003)</math>, so <math>a</math> has <math>\boxed{501}</math> possible values. | ||
Line 15: | Line 18: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:35, 1 March 2020
Contents
Problem
Positive integers ,
,
, and
satisfy
,
, and
. Find the number of possible values of
.
Solution 1
Using the difference of squares, , where equality must hold so
and
. Then we see
is maximal and
is minimal, so the answer is
.
Note: We can also find that in another way. We know
Therefore, one of and
must be
Clearly,
since then one would be positive and negative, or both would be zero. Therefore,
so
. Similarly, we can deduce that
Solution 2
Since must be greater than
, it follows that the only possible value for
is
(otherwise the quantity
would be greater than
). Therefore the only possible ordered pairs for
are
,
, ... ,
, so
has
possible values.
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.