BDMO National Junior 2018/9

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nahin munkar
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BDMO National Junior 2018/9

Unread post by nahin munkar » Tue Jan 08, 2019 2:01 pm

Find the number of positive integers that are divisors of at least one of $10^{10}$, $12^{12}$, $15^{15}$.
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samiul_samin
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Re: BDMO National Junior 2018/9

Unread post by samiul_samin » Fri Jan 11, 2019 12:09 pm

Answer
$671$
$Sol^n$
Let the divisors of,
$10^{10}=a$
$12^{12}=b$
$15^{15}=c$
So,
$a=11×11=121$
$b=25×13=325$
$c=16×16=256$
Common numbers of ,
a&c =$10$
b&c=$12$
c&a=$10$
a&b&c=$1$

So,answer=$121+325+256-10-12-10+1=671$

Using PIE

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