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}\\ | ||
| − | &= a^4 + 2 \cdot a^ | + | &= 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> | ||
| − | (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). | + | (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, | ||
Latest revision as of 11:08, 10 March 2012
Problem 12
Let 
and 
. If 
, how many integral values of 
are there such that 
is a multiple of 
?
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*}](http://latex.artofproblemsolving.com/c/e/a/cea171bfac003b9497fc76bf9525eb81e7d289cd.png)
(If you recall the reverse of Sophie Germain Identity with 
, then you could have directly found the answer).
By Fermat's Little Theorem, we have that 
if 
and 
if 
. Also, we note that by examining a couple of terms, 
if 
and 
if 
. Therefore,
![\[a^{4} + 2^{2a} \equiv \{0,1\} + \{1,4\} \equiv \{0,1,2,4\} \pmod{5}\]](http://latex.artofproblemsolving.com/e/f/1/ef103f9a442d11c72c54040783eee9ef5aa5bb04.png)
With divisibility by achievable only if 
. There are 
odd numbers in the range given, and 
of those are divisible by , so the answer is 
.
See also
| Mock AIME 1 2007-2008 (Problems, Source) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
