Difference between revisions of "2011 AIME I Problems/Problem 15"
(Created page with '== Problem == For some integer <math>m</math>, the polynomial <math>x^3 - 2011x + m</math> has the three integer roots <math>a</math>, <math>b</math>, and <math>c</math>. Find <…') |
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<tr><th><math>a</math></th><th><math>a^2 = 2011 + bc</math></th></tr> | <tr><th><math>a</math></th><th><math>a^2 = 2011 + bc</math></th></tr> | ||
− | <tr><td><math>45</math></td><td><math>14 | + | <tr><td><math>45</math></td><td><math>14</math></td></tr> |
<tr><td><math>46</math></td><td><math>14 + 91 =105</math></td></tr> | <tr><td><math>46</math></td><td><math>14 + 91 =105</math></td></tr> | ||
<tr><td><math>47</math></td><td><math>105 + 93 = 198</math></td></tr> | <tr><td><math>47</math></td><td><math>105 + 93 = 198</math></td></tr> | ||
Line 54: | Line 54: | ||
<math>105 = (3)(5)(7)</math>, So <math>46</math> won't work. <math>198/47 > 4</math>, | <math>105 = (3)(5)(7)</math>, So <math>46</math> won't work. <math>198/47 > 4</math>, | ||
− | <math>198 is not divisible by 5</math>, <math>198/6 = 33</math>, which is too small to get <math>47</math> | + | <math>198</math> is not divisible by <math>5</math>, <math>198/6 = 33</math>, which is too small to get <math>47</math> |
<math>293/48 > 6</math>, <math>293</math> is not divisible by <math>7</math> or <math>8</math> or <math>9</math>, we can clearly tell that <math>10</math> is too much | <math>293/48 > 6</math>, <math>293</math> is not divisible by <math>7</math> or <math>8</math> or <math>9</math>, we can clearly tell that <math>10</math> is too much | ||
+ | |||
Hence, <math>a = 49</math>, <math>a^2 -2011 = 390</math>. <math>b = 39</math>, <math>c = 10</math>. | Hence, <math>a = 49</math>, <math>a^2 -2011 = 390</math>. <math>b = 39</math>, <math>c = 10</math>. | ||
Answer: <math>098</math> | Answer: <math>098</math> | ||
+ | |||
+ | == See also == | ||
+ | |||
+ | {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 23:09, 19 March 2011
Problem
For some integer , the polynomial has the three integer roots , , and . Find .
Solution
With Vieta's formula, we know that , and .
since any one being zero will make the the other 2 .
. WLOG, let .
Then if , then and if , .
We know that , have the same sign. So . ( and )
Also, maximize when if we fixed . Hence, .
So .
so .
Now we have limited a to .
Let's us analyze .
Here is a table:
We can tell we don't need to bother with ,
, So won't work. ,
is not divisible by , , which is too small to get
, is not divisible by or or , we can clearly tell that is too much
Hence, , . , .
Answer:
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |