Integers a, b, c, d, and e satisfy the following three properties: (i) $2 \le a < b <c <d <e <100$ (ii)$ \gcd (a,e) = 1 $ (iii) a, b, c, d, e form a geometric sequence. What is the value of c?
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sicilianfan
04.11.2012 15:52
Rewriting $a,b,c,d,e$ as $a,ar,ar^2,ar^3,ar^4$, it can be seen that the $gcd(a,e)$ is $a$. Therefore, $a$ must be equal to $1$. However in the problem, it states that $a \ge 2$ so it is not possible for $a$ to equal $1$. Therefore, I believe that this question is flawed.
kevinnoh
04.11.2012 16:52
i think $a=k^{4}$ for some integer $k$ for there to be a solution because if you make $r$ a fraction with a denominator that is $k$ then you could eliminate all their common factors and have $\gcd(a,e)=1$. after very little inspection... you can see that $a=16$ is the only case that could possibly work and thus follows $r=\frac{3}{2}$ which would make $\boxed{c=36}$
10000th User
04.11.2012 17:04
sicilianfan wrote:
Rewriting $a,b,c,d,e$ as $a,ar,ar^2,ar^3,ar^4$, it can be seen that the $gcd(a,e)$ is $a$. Therefore, $a$ must be equal to $1$. However in the problem, it states that $a \ge 2$ so it is not possible for $a$ to equal $1$. Therefore, I believe that this question is flawed.
Your assumption about $r$ being an integer is flawed. $\gcd(a,e)=a$ if $r$ is a nonzero integer