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Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 12"

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== Solution ==
 
== Solution ==
 
<cmath>\begin{align*}[(a^2 + 2^a) + a \cdot 2^{(a+1)/2}][(a^2 + 2^a) - a \cdot 2^{(a+1)/2}] &= (a^2 + 2^a)^2 - a^2 \cdot 2^{a+1}\\
 
<cmath>\begin{align*}[(a^2 + 2^a) + a \cdot 2^{(a+1)/2}][(a^2 + 2^a) - a \cdot 2^{(a+1)/2}] &= (a^2 + 2^a)^2 - a^2 \cdot 2^{a+1}\\
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&= a^4 + 2 \cdot a^42^{a} + 2^{2a} - a^2 \cdot 2^{a+1}\\
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&= a^4 + 2 \cdot a^22^{a} + 2^{2a} - a^2 \cdot 2^{a+1}\\
 
&= a^4 + 2^{2a}\end{align*}</cmath>
 
&= a^4 + 2^{2a}\end{align*}</cmath>
  
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(If you recall the reverse of [[Sophie Germain Identity]] with <math>a=a,\, b = 2^{a/2}</math>, then you could have directly found the answer).
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(If you recall the reverse of [[Sophie Germain Identity]] with <math>a=a,\, b = 2^{(a-1)/2}</math>, then you could have directly found the answer).
  
 
By [[Fermat's Little Theorem]], we have that <math>a^{4} \equiv 1 \pmod{5}</math> if <math>a \nmid 5</math> and <math>a^{4} \equiv 0 \pmod{5}</math> if <math>a | 5</math>. Also, we note that by examining a couple of terms, <math>2^{2a} \equiv 4 \pmod{5}</math> if <math>a \nmid 2</math> and <math>2^{2a} \equiv 1 \pmod{5}</math> if <math>a|2</math>. Therefore,  
 
By [[Fermat's Little Theorem]], we have that <math>a^{4} \equiv 1 \pmod{5}</math> if <math>a \nmid 5</math> and <math>a^{4} \equiv 0 \pmod{5}</math> if <math>a | 5</math>. Also, we note that by examining a couple of terms, <math>2^{2a} \equiv 4 \pmod{5}</math> if <math>a \nmid 2</math> and <math>2^{2a} \equiv 1 \pmod{5}</math> if <math>a|2</math>. Therefore,  
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Number Theory Problems]]

Latest revision as of 11:08, 10 March 2012

Problem 12

Let $d_1 = a^2 + 2^a + a \cdot 2^{(a+1)/2}$ and $d_2 = a^2 + 2^a - a \cdot 2^{(a+1)/2}$. If $1 \le a \le 251$, how many integral values of $a$ are there such that $d_1 \cdot d_2$ is a multiple of $5$?

Solution

\begin{align*}[(a^2 + 2^a) + a \cdot 2^{(a+1)/2}][(a^2 + 2^a) - a \cdot 2^{(a+1)/2}] &= (a^2 + 2^a)^2 - a^2 \cdot 2^{a+1}\\ &= a^4 + 2 \cdot a^22^{a} + 2^{2a} - a^2 \cdot 2^{a+1}\\ &= a^4 + 2^{2a}\end{align*}

(If you recall the reverse of Sophie Germain Identity with $a=a,\, b = 2^{(a-1)/2}$, then you could have directly found the answer).

By Fermat's Little Theorem, we have that $a^{4} \equiv 1 \pmod{5}$ if $a \nmid 5$ and $a^{4} \equiv 0 \pmod{5}$ if $a | 5$. Also, we note that by examining a couple of terms, $2^{2a} \equiv 4 \pmod{5}$ if $a \nmid 2$ and $2^{2a} \equiv 1 \pmod{5}$ if $a|2$. Therefore, \[a^{4} + 2^{2a} \equiv \{0,1\} + \{1,4\} \equiv \{0,1,2,4\} \pmod{5}\] With divisibility by $5$ achievable only if $a \nmid 2,5$. There are $\frac{251-1}{2}+1 = 126$ odd numbers in the range given, and $\frac{245-5}{10}+1 = 25$ of those are divisible by $5$, so the answer is $126 - 25 = \boxed{101}$.

See also

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by
Problem 11 		Followed by
Problem 13
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