Difference between revisions of "2023 AIME I Problems/Problem 9"
(→Solution 1) |
m (→Solution 1) |
||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
===Solution 1=== | ===Solution 1=== | ||
− | |||
Plugging <math>2</math> into <math>P(x)</math>, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. We can rewrite into <math>(2-m)(m^2+2m+4+a(2+m)+b)=0</math>, where <math>c</math> can be any value in the range. Since <math>m\neq2, m^2+2m+4+a(2+m)+b</math> must be <math>0</math>. The problem also asks for unique integers, meaning <math>m</math> can only be one value for each polynomial, so the discriminant must be <math>0</math>. <math>m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0</math>, and <math>(2+a)^2-4(2a+b+4)=0</math>. Rewrite to be <math>a(a-4)=4(b+3)</math>. <math>a</math> must be even for <math>4(b+3)</math> to be an integer. <math>-10<=a<=10</math> because <math>4(20+3) = 92</math>. There are 11 pairs of <math>(a,b)</math> and 41 integers for <math>c</math>, giving <cmath>41\cdot11 = \boxed{451}</cmath> | Plugging <math>2</math> into <math>P(x)</math>, we get <math>8+4a+2b+c = m^3+am^2+bm+c</math>. We can rewrite into <math>(2-m)(m^2+2m+4+a(2+m)+b)=0</math>, where <math>c</math> can be any value in the range. Since <math>m\neq2, m^2+2m+4+a(2+m)+b</math> must be <math>0</math>. The problem also asks for unique integers, meaning <math>m</math> can only be one value for each polynomial, so the discriminant must be <math>0</math>. <math>m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0</math>, and <math>(2+a)^2-4(2a+b+4)=0</math>. Rewrite to be <math>a(a-4)=4(b+3)</math>. <math>a</math> must be even for <math>4(b+3)</math> to be an integer. <math>-10<=a<=10</math> because <math>4(20+3) = 92</math>. There are 11 pairs of <math>(a,b)</math> and 41 integers for <math>c</math>, giving <cmath>41\cdot11 = \boxed{451}</cmath> |
Revision as of 14:11, 8 February 2023
Contents
[hide]Problem (Unofficial, please update when official one comes out):
is a polynomial with integer coefficients in the range, inclusive. There is exactly one integer such that . How many possible values are there for the ordered triple ?
Solution
Solution 1
Plugging into , we get . We can rewrite into , where can be any value in the range. Since must be . The problem also asks for unique integers, meaning can only be one value for each polynomial, so the discriminant must be . , and . Rewrite to be . must be even for to be an integer. because . There are 11 pairs of and 41 integers for , giving
~chem1kall