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Difference between revisions of "2015 AIME I Problems/Problem 13"

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With all angles measured in degrees, the product <math>\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n</math>, where <math>m</math> and <math>n</math> are integers greater than 1. Find <math>m+n</math>.
 
With all angles measured in degrees, the product <math>\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n</math>, where <math>m</math> and <math>n</math> are integers greater than 1. Find <math>m+n</math>.
  
−
==Solution==
+
==Solution 1==
−
===Solution 1===
 
 
Let <math>x = \cos 1^\circ + i \sin 1^\circ</math>. Then from the identity
 
Let <math>x = \cos 1^\circ + i \sin 1^\circ</math>. Then from the identity
 
<cmath>\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},</cmath>
 
<cmath>\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},</cmath>
Line 15: Line 14:
 
It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{91}</math>.
 
It is easy to see that <math>M = 2^{89}</math> and that our answer is <math>2 + 89 = \boxed{91}</math>.
  
−
===Solution 2===
+
==Solution 2==
 
Let <math>p=\sin1\sin3\sin5...\sin89</math>
 
Let <math>p=\sin1\sin3\sin5...\sin89</math>
  
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Thus the answer is <math>2+89=\boxed{091}</math>
 
Thus the answer is <math>2+89=\boxed{091}</math>
  
−
=== Solution 3 ===
+
== Solution 3 ==
 
Similar to Solution <math>2</math>, so we use <math>\sin{2\theta}=2\sin\theta\cos\theta</math> and we find that:
 
Similar to Solution <math>2</math>, so we use <math>\sin{2\theta}=2\sin\theta\cos\theta</math> and we find that:
 
<cmath>\begin{align*}\sin(4)\sin(8)\sin(12)\sin(16)\cdots\sin(84)\sin(88)&=(2\sin(2)\cos(2))(2\sin(4)\cos(4))(2\sin(6)\cos(6))(2\sin(8)\cos(8))\cdots(2\sin(42)\cos(42))(2\sin(44)\cos(44))\\
 
<cmath>\begin{align*}\sin(4)\sin(8)\sin(12)\sin(16)\cdots\sin(84)\sin(88)&=(2\sin(2)\cos(2))(2\sin(4)\cos(4))(2\sin(6)\cos(6))(2\sin(8)\cos(8))\cdots(2\sin(42)\cos(42))(2\sin(44)\cos(44))\\
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The answer is therefore <math>m+n=(2)+(89)=\boxed{091}</math>.
 
The answer is therefore <math>m+n=(2)+(89)=\boxed{091}</math>.
  
−
===Solution 4===
+
==Solution 4==
 
Let <math>p=\prod_{k=1}^{45} \csc^2(2k-1)^\circ</math>.
 
Let <math>p=\prod_{k=1}^{45} \csc^2(2k-1)^\circ</math>.
  
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Using the fact that <math>\sin\theta=\cos(90^{\circ}-\theta)</math> again, our equation simplifies to <math>\sqrt{\frac{1}{2p}}=\frac{\sin90^\circ}{2^{45}}</math>, and since <math>\sin90^\circ=1</math>, it follows that <math>2p = 2^{90}</math>, which implies <math>p=2^{89}</math>. Thus, <math>m+n=2+89=\boxed{091}</math>.
 
Using the fact that <math>\sin\theta=\cos(90^{\circ}-\theta)</math> again, our equation simplifies to <math>\sqrt{\frac{1}{2p}}=\frac{\sin90^\circ}{2^{45}}</math>, and since <math>\sin90^\circ=1</math>, it follows that <math>2p = 2^{90}</math>, which implies <math>p=2^{89}</math>. Thus, <math>m+n=2+89=\boxed{091}</math>.
−
===Solution 5===
+
==Solution 5==
 
Once we have the tools of complex polynomials there is no need to use the tactical tricks. Everything is so basic (I think).
 
Once we have the tools of complex polynomials there is no need to use the tactical tricks. Everything is so basic (I think).
  
Line 87: Line 86:
 
<cmath>\prod_{k=1}^{45}\csc^2\frac{(2k-1)\pi}{180} = 2^{89}</cmath>
 
<cmath>\prod_{k=1}^{45}\csc^2\frac{(2k-1)\pi}{180} = 2^{89}</cmath>
 
-Mathdummy
 
-Mathdummy
 +
==Solution 6==
 +
Recall that <math>\sin\alpha\cdot \sin(60^{\circ}-\alpha)\cdot \sin(60^{\circ}+\alpha)=\frac{1}{4}\cdot \sin3\alpha</math>
 +
Since it is in csc, we can write in sin and then take reciprocal.
 +
We can group them by threes, <math>P=(\sin1^{\circ}\cdot \sin59^{\circ}\cdot \sin61^{\circ})\cdots(\sin29^{\circ}\cdot \sin31^{\circ}\cdot \sin89^{\circ})</math>. Thus
 +
<cmath>\begin{align*}
 +
P &=\frac{1}{4^{15}}\cdot \sin3^{\circ}\cdot \sin9^{\circ}\cdots\sin87^{\circ}\\
 +
&=\frac{1}{4^{20}}\cdot \sin9^{\circ}\cdot \sin27^{\circ}\cdot \sin45^{\circ}\cdot \sin63^{\circ}\cdot \sin81^{\circ}\\
 +
&=\frac{1}{4^{20}}\cdot \frac{\sqrt{2}}{2}\cdot \sin9^{\circ}\cdot \cos9^{\circ}\cdot \sin27^{\circ}\cdot \cos27^{\circ}\\
 +
&=\frac{1}{4^{21}}\cdot \frac{\sqrt{2}}{2}\cdot \sin18^{\circ}\cdot \cos36^{\circ}=\frac{\sqrt{2}}{2^{45}}
 +
\end{align*}</cmath>
 +
So we take reciprocal, <math>\frac 1P=2^{\frac{89}{2}}</math>, the desired answer is <math>\frac{1}{P^2}=2^{89}</math> leads to answer <math>\boxed{091}</math>
 +
 +
~bluesoul
 +
 +
==Solution 7==
 +
 +
We have
 +
 +
<cmath>\prod_{k=1}^{45} \csc^2(2k-1)^\circ = \left(\frac{1}{\sin1^\circ \cdot \sin3^\circ \cdots \sin89^\circ}\right)^2.</cmath>
 +
 +
Multiplying by <math>\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}</math> gives
 +
 +
<cmath>\left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin88^\circ \cdot \sin89^\circ}\right)^2</cmath>
 +
 +
<cmath> = \left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin 45^\circ \cdot \cos 44^\circ \cdot \cos 43^\circ \cdots \cos1^\circ}\right)^2.</cmath>
 +
 +
Using <math>\sin\alpha \cos\alpha = \frac{1}{2}\sin{2\alpha}</math> gives
 +
 +
<cmath> \left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\frac{1}{2} \sin2^\circ \cdot \frac{1}{2} \sin4^\circ \cdots \frac{1}{2} \sin88^\circ \cdot \sin45^\circ}\right) ^2 </cmath>
 +
 +
<cmath> = \left(\frac{1}{(\frac{1}{2})^{44} \cdot \frac{\sqrt{2}}{2}}\right)^2 </cmath>
 +
 +
<cmath> = 2^{89}. </cmath>
 +
 +
Thus, the answer is <math>2+89 = \boxed{091}.</math>
 +
 +
==Solution 8==
 +
Consider the product <math>\prod_{k=1}^{45} \csc^2(2k-1) = \prod_{k=45}^{1} \sec^2(2k-1) = \prod_{k=1}^{45} \sec^2(2k-1) =  \prod_{k=1}^{45} (1+\tan^2(2k-1))</math>
 +
However, note that the <math>45</math> numbers in the form <math>\sqrt{\tan^2(2k-1)} = \tan(2k-1)</math> for <math>1\le{k}\le{45}</math> are precisely the roots of the equation <math>\frac{1}{\tan{(\tan^{-1}{90x}})} = 0</math>. Thus, it suffices to find <math>|P(-1)|</math>, where <math>P</math> is the polynomial formed by the denominator of <math>\tan{(\tan^{-1}{90\sqrt{x}})}</math>. This is true because <math>\prod_{k=1}^{45} (x-\tan^2(2k-1)) = \prod_{k=1}^{45} -(-x+\tan^2(2k-1))</math> gives us the factored root form of the equation where <math>P</math> is undefined, which corresponds to the equation where the denominator equals <math>0</math>.
 +
 +
It remains to find the denominator of <math>P</math>; fortunately, we may use tangent angle multiplication rules. Specifically, the denominator of <math>P</math> will be <math>\sum_{n=0} ^{45} (-1)^n\binom{90}{2n}\sqrt{x}^{2n} = \sum_{n=0} ^{45} (-1)^n\binom{90}{2n}x^n</math>. Evaluating at <math>x = -1</math>, we may obtain the sum <math>\sum_{n=0} ^{45} (-1)^n\binom{90}{2n}(-1)^{n} = \sum_{n=0} ^{45} \binom{90}{2n}</math>.
 +
 +
From here, there are two ways to finish; the first is to recognize the well known sum <math>\sum_{n=0} ^{k} \binom{2k}{2n} = 2^(2k-1)</math>, from which we may plug in <math>k = 45</math> to see <math>|P(-1)| = 2^(45*2-1) = 2^{89}</math> to obtain the answer of <math>2+89=\boxed{091}</math>. Otherwise, you may split the previous sum; <math>\sum_{n=0} ^{45} \binom{90}{2n} = \sum_{n=0} ^{45} \binom{89}{2n-1} + \binom{89}{2n} = \sum_{n=0} ^{89} \binom{89}{n} = 2^{89}</math>, which also gives the correct answer. <math>\blacksquare</math>
 +
 +
 +
 +
  
  

Latest revision as of 19:51, 18 February 2025

Problem

With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$, where $m$ and $n$ are integers greater than 1. Find $m+n$.

Solution 1

Let $x = \cos 1^\circ + i \sin 1^\circ$. Then from the identity \[\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},\] we deduce that (taking absolute values and noticing $|x| = 1$) \[|2\sin 1| = |x^2 - 1|.\] But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$, if we let our product be $M$ then \[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ\] \[= \frac{1}{2^{90}} |x^2 - 1| |x^6 - 1| |x^{10} - 1| \dots |x^{354} - 1| |x^{358} - 1|\] because $\sin$ is positive in the first and second quadrants. Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z - x^2)(z - x^6)\dots (z - x^{358}) = z^{90} + 1$, and so \[\frac{1}{M} = \dfrac{1}{2^{90}}|1 - x^2| |1 - x^6| \dots |1 - x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.\] It is easy to see that $M = 2^{89}$ and that our answer is $2 + 89 = \boxed{91}$.

Solution 2

Let $p=\sin1\sin3\sin5...\sin89$

\[p=\sqrt{\sin1\sin3\sin5...\sin177\sin179}\]

\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}\]

\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{(2\sin1\cos1)\cdot(2\sin2\cos2)\cdot(2\sin3\cos3)\cdot....\cdot(2\sin89\cos89)}}\]

\[=\sqrt{\frac{1}{2^{89}}\frac{\sin90\sin91\sin92\sin93...\sin177\sin178\sin179}{\cos1\cos2\cos3\cos4...\cos89}}\]

$=\sqrt{\frac{1}{2^{89}}}$ because of the identity $\sin(90+x)=\cos(x)$

we want $\frac{1}{p^2}=2^{89}$

Thus the answer is $2+89=\boxed{091}$

Solution 3

Similar to Solution $2$, so we use $\sin{2\theta}=2\sin\theta\cos\theta$ and we find that: \begin{align*}\sin(4)\sin(8)\sin(12)\sin(16)\cdots\sin(84)\sin(88)&=(2\sin(2)\cos(2))(2\sin(4)\cos(4))(2\sin(6)\cos(6))(2\sin(8)\cos(8))\cdots(2\sin(42)\cos(42))(2\sin(44)\cos(44))\\ &=(2\sin(2)\sin(88))(2\sin(4))\sin(86))(2\sin(6)\sin(84))(2\sin(8)\sin(82))\cdots(2\sin(42)\sin(48))(2\sin(44)\sin(46))\\ &=2^{22}(\sin(2)\sin(88)\sin(4)\sin(86)\sin(6)\sin(84)\sin(8)\sin(82)\cdots\sin(42)\sin(48)\sin(44)\sin(46))\\ &=2^{22}(\sin(2)\sin(4)\sin(6)\sin(8)\cdots\sin(82)\sin(84)\sin(86)\sin(88))\end{align*} Now we can cancel the sines of the multiples of $4$: \[1=2^{22}(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86))\] So $\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)=2^{-22}$ and we can apply the double-angle formula again: \begin{align*}2^{-22}&=(\sin(2)\sin(6)\sin(10)\sin(14)\cdots\sin(82)\sin(86)\\ &=(2\sin(1)\cos(1))(2\sin(3)\cos(3))(2\sin(5)\cos(5))(2\sin(7)\cos(7))\cdots(2\sin(41)\cos(41))(2\sin(43)\cos(43))\\ &=(2\sin(1)\sin(89))(2\sin(3)\sin(87))(2\sin(5)\sin(85))(2\sin(7)\sin(87))\cdots(2\sin(41)\sin(49))(2\sin(43)\sin(47))\\ &=2^{22}(\sin(1)\sin(89)\sin(3)\sin(87)\sin(5)\sin(85)\sin(7)\sin(83)\cdots\sin(41)\sin(49)\sin(43)\sin(47))\\ &=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\end{align*} Of course, $\sin(45)=2^{-\frac{1}{2}}$ is missing, so we multiply it to both sides: \[2^{-22}\sin(45)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(45))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] \[\left(2^{-22}\right)\left(2^{-\frac{1}{2}}\right)=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] \[2^{-\frac{45}{2}}=2^{22}(\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\] Now isolate the product of the sines: \[\sin(1)\sin(3)\sin(5)\sin(7)\cdots\sin(83)\sin(85)\sin(87)\sin(89)=2^{-\frac{89}{2}}\] And the product of the squares of the cosecants as asked for by the problem is the square of the inverse of this number: \[\csc^2(1)\csc^2(3)\csc^2(5)\csc^2(7)\cdots\csc^2(83)\csc^2(85)\csc^2(87)\csc^2(89)=\left(\frac{1}{2^{-\frac{89}{2}}}\right)^2=\left(2^{\frac{89}{2}}\right)^2=2^{89}\] The answer is therefore $m+n=(2)+(89)=\boxed{091}$.

Solution 4

Let $p=\prod_{k=1}^{45} \csc^2(2k-1)^\circ$.

Then, $\sqrt{\frac{1}{p}}=\prod_{k=1}^{45} \sin(2k-1)^\circ$.

Since $\sin\theta=\cos(90^{\circ}-\theta)$, we can multiply both sides by $\frac{\sqrt{2}}{2}$ to get $\sqrt{\frac{1}{2p}}=\prod_{k=1}^{23} \sin(2k-1)^\circ\cos(2k-1)^\circ$.

Using the double-angle identity $\sin2\theta=2\sin\theta\cos\theta$, we get $\sqrt{\frac{1}{2p}}=\frac{1}{2^{23}}\prod_{k=1}^{23} \sin(4k-2)^\circ$.

Note that the right-hand side is equal to $\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \sin(4k)^\circ$, which is equal to $\frac{1}{2^{23}}\prod_{k=1}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} 2\sin(2k)^\circ\cos(2k)^\circ$, again, from using our double-angle identity.

Putting this back into our equation and simplifying gives us $\sqrt{\frac{1}{2p}}=\frac{1}{2^{45}}\prod_{k=23}^{45} \sin(2k)^\circ\div \prod_{k=1}^{22} \cos(2k)^\circ$.

Using the fact that $\sin\theta=\cos(90^{\circ}-\theta)$ again, our equation simplifies to $\sqrt{\frac{1}{2p}}=\frac{\sin90^\circ}{2^{45}}$, and since $\sin90^\circ=1$, it follows that $2p = 2^{90}$, which implies $p=2^{89}$. Thus, $m+n=2+89=\boxed{091}$.

Solution 5

Once we have the tools of complex polynomials there is no need to use the tactical tricks. Everything is so basic (I think).

Recall that the roots of $x^n+1$ are $e^{\frac{(2k-1)\pi i}{n}}, k=1,2,...,n$, we have \[x^n + 1 = \prod_{k=1}^{n}(x-e^{\frac{(2k-1)\pi i}{n}})\] Let $x=1$, and take absolute value of both sides, \[2 = \prod_{k=1}^{n}|1-e^{\frac{(2k-1)\pi i}{n}}|= 2^n\prod_{k=1}^{n}|\sin\frac{(2k-1)\pi}{2n}|\] or, \[\prod_{k=1}^{n}|\sin\frac{(2k-1)\pi}{2n}| = 2^{-(n-1)}\] Let $n$ be even, then, \[\sin\frac{(2k-1)\pi}{2n} = \sin\left(\pi - \frac{(2k-1)\pi}{2n}\right) = \sin\left(\frac{(2(n-k+1)-1)\pi}{2n}\right)\] so, \[\prod_{k=1}^{n}\left|\sin\frac{(2k-1)\pi}{n}\right| = \prod_{k=1}^{\frac{n}{2}}\sin^2\frac{(2k-1)\pi}{2n}\] Set $n=90$ and we have \[\prod_{k=1}^{45}\sin^2\frac{(2k-1)\pi}{180} = 2^{-89}\], \[\prod_{k=1}^{45}\csc^2\frac{(2k-1)\pi}{180} = 2^{89}\] -Mathdummy

Solution 6

Recall that $\sin\alpha\cdot \sin(60^{\circ}-\alpha)\cdot \sin(60^{\circ}+\alpha)=\frac{1}{4}\cdot \sin3\alpha$ Since it is in csc, we can write in sin and then take reciprocal. We can group them by threes, $P=(\sin1^{\circ}\cdot \sin59^{\circ}\cdot \sin61^{\circ})\cdots(\sin29^{\circ}\cdot \sin31^{\circ}\cdot \sin89^{\circ})$. Thus \begin{align*} P &=\frac{1}{4^{15}}\cdot \sin3^{\circ}\cdot \sin9^{\circ}\cdots\sin87^{\circ}\\ &=\frac{1}{4^{20}}\cdot \sin9^{\circ}\cdot \sin27^{\circ}\cdot \sin45^{\circ}\cdot \sin63^{\circ}\cdot \sin81^{\circ}\\ &=\frac{1}{4^{20}}\cdot \frac{\sqrt{2}}{2}\cdot \sin9^{\circ}\cdot \cos9^{\circ}\cdot \sin27^{\circ}\cdot \cos27^{\circ}\\ &=\frac{1}{4^{21}}\cdot \frac{\sqrt{2}}{2}\cdot \sin18^{\circ}\cdot \cos36^{\circ}=\frac{\sqrt{2}}{2^{45}} \end{align*} So we take reciprocal, $\frac 1P=2^{\frac{89}{2}}$, the desired answer is $\frac{1}{P^2}=2^{89}$ leads to answer $\boxed{091}$

~bluesoul

Solution 7

We have

\[\prod_{k=1}^{45} \csc^2(2k-1)^\circ = \left(\frac{1}{\sin1^\circ \cdot \sin3^\circ \cdots \sin89^\circ}\right)^2.\]

Multiplying by $\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}$ gives

\[\left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin88^\circ \cdot \sin89^\circ}\right)^2\]

\[= \left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\sin1^\circ \sin2^\circ \cdot \sin3^\circ \cdots \sin 45^\circ \cdot \cos 44^\circ \cdot \cos 43^\circ \cdots \cos1^\circ}\right)^2.\]

Using $\sin\alpha \cos\alpha = \frac{1}{2}\sin{2\alpha}$ gives

\[\left(\frac{\sin2^\circ \cdot \sin4^\circ \cdots \sin88^\circ}{\frac{1}{2} \sin2^\circ \cdot \frac{1}{2} \sin4^\circ \cdots \frac{1}{2} \sin88^\circ \cdot \sin45^\circ}\right) ^2\]

\[= \left(\frac{1}{(\frac{1}{2})^{44} \cdot \frac{\sqrt{2}}{2}}\right)^2\]

\[= 2^{89}.\]

Thus, the answer is $2+89 = \boxed{091}.$

Solution 8

Consider the product $\prod_{k=1}^{45} \csc^2(2k-1) = \prod_{k=45}^{1} \sec^2(2k-1) = \prod_{k=1}^{45} \sec^2(2k-1) = \prod_{k=1}^{45} (1+\tan^2(2k-1))$ However, note that the $45$ numbers in the form $\sqrt{\tan^2(2k-1)} = \tan(2k-1)$ for $1\le{k}\le{45}$ are precisely the roots of the equation $\frac{1}{\tan{(\tan^{-1}{90x}})} = 0$. Thus, it suffices to find $|P(-1)|$, where $P$ is the polynomial formed by the denominator of $\tan{(\tan^{-1}{90\sqrt{x}})}$. This is true because $\prod_{k=1}^{45} (x-\tan^2(2k-1)) = \prod_{k=1}^{45} -(-x+\tan^2(2k-1))$ gives us the factored root form of the equation where $P$ is undefined, which corresponds to the equation where the denominator equals $0$.

It remains to find the denominator of $P$; fortunately, we may use tangent angle multiplication rules. Specifically, the denominator of $P$ will be $\sum_{n=0} ^{45} (-1)^n\binom{90}{2n}\sqrt{x}^{2n} = \sum_{n=0} ^{45} (-1)^n\binom{90}{2n}x^n$. Evaluating at $x = -1$, we may obtain the sum $\sum_{n=0} ^{45} (-1)^n\binom{90}{2n}(-1)^{n} = \sum_{n=0} ^{45} \binom{90}{2n}$.

From here, there are two ways to finish; the first is to recognize the well known sum $\sum_{n=0} ^{k} \binom{2k}{2n} = 2^(2k-1)$, from which we may plug in $k = 45$ to see $|P(-1)| = 2^(45*2-1) = 2^{89}$ to obtain the answer of $2+89=\boxed{091}$. Otherwise, you may split the previous sum; $\sum_{n=0} ^{45} \binom{90}{2n} = \sum_{n=0} ^{45} \binom{89}{2n-1} + \binom{89}{2n} = \sum_{n=0} ^{89} \binom{89}{n} = 2^{89}$, which also gives the correct answer. $\blacksquare$




See Also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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