MHB Finding $a_{50}$ with Coprime Constraints

[Albert1]

given :
$a_1<a_2<a_3<--------<a_{50}$
$a_1,a_2,a_3,------,a_{50}\in N$
all $a_1,a_2,------,a_{50}$ are coprime with 987
that is $(a_n,987)=1 $
here $1\leq n\leq 50$
please find $a_{50}$

 

[Nono713]

Spoiler

I think you need to add a little more information, else the challenge is trivial - is $a_n$ intended to be the $n$th smallest integer coprime with 987? As described right now, $a_n = 987n + 1$ satisfies all your conditions.

 

[Albert1]

Spoiler

I think you need to add a little more information, else the challenge is trivial - is $a_n$ intended to be the $n$th smallest integer coprime with 987? As described right now, $a_n = 987n + 1$ satisfies all your conditions.

yes ,$a_n$ is the $n$th smallest integer coprime with 987
and please find $a_{50}$

 

[kaliprasad]

Spoiler

First we need to factor 987
987 = 3 * 329 = 3 * 7 * 47
to find a number which is coprime to 987 it should be co-prime to 3,7, and 47
now 1 is not coprime to any number so we need to find the 51st number which is not divisible by 3 7 or 47
for x the numbers below or same as not divisible by 3 7 and 47 are
$f(x) =x-\lfloor\dfrac{x}{3}\rfloor-\lfloor\dfrac{x}{7}\rfloor-\lfloor\dfrac{x}{47}\rfloor+\lfloor\dfrac{x}{3*7}\rfloor+\lfloor\dfrac{x}{3* 47}\rfloor+\lfloor\dfrac{x}{7*47}\rfloor- \lfloor\dfrac{x}{3 * 7 * 47}\rfloor$
or $f(x) =x-\lfloor\dfrac{x}{3}\rfloor-\lfloor\dfrac{x}{7}\rfloor-\lfloor\dfrac{x}{47}\rfloor+\lfloor\dfrac{x}{21}\rfloor+\lfloor\dfrac{x}{141}\rfloor+\lfloor\dfrac{x}{329}\rfloor- \lfloor\dfrac{x}{987}\rfloor$
or $f(x) =x-\lfloor\dfrac{x}{3}\rfloor-\lfloor\dfrac{x}{7}\rfloor+\lfloor\dfrac{x}{21}\rfloor-\lfloor\dfrac{x}{47}\rfloor+\lfloor\dfrac{x}{141}\rfloor+\lfloor\dfrac{x}{329}\rfloor- \lfloor\dfrac{x}{987}\rfloor$
for estimating we take $x-\dfrac{x}{3}=\dfrac{2x}{3}=51$ or x = 76 ( rounded)
so f (x) = 76 - 25 - 10 + 3 = 44
we are falling short by 7
so we add 14 as it is > 7 * 3/2
so we get x = 90 but as 90 is not coprime we take 89
f(89) = 89 - 29 - 12 + 4 - 1 = 51
so x = 89 is the ans.

 

[Albert1]

$a_1=1$ should be included $a_{50}=88$

 

[kaliprasad]

$a_1=1$ should be included $a_{50}=88$

That is slight debatable as 1 is not coprime to any number so 1 should be left out. But if you want to consider 1 then you are right.

 

[Albert1]

Spoiler

$987=3\times 7\times 47$
let :$a_{50}=x$
we have :
$\big[\dfrac{x}{3}\big]+[\dfrac{x}{7}]+[\dfrac{x}{47}]-[\dfrac{x}{21}]=x-50=38$
and we get :$ x=a_{50}=88$