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Find all positive integers $a$ and $b$ satisfying $$\gcd (a,b)=10$$ and $$\operatorname{lcm} (a,b)=100$$ simultaneously.

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closed as off-topic by Adam Chalumeau, Ak19, lulu, John Omielan, max_zorn Jul 29 at 2:33

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  • 1
    $\begingroup$ Welcome to MSE. Please show us what you've tried. $\endgroup$ – Ak19 Jul 28 at 12:07
  • $\begingroup$ $100$ only has nine divisors....if you can't think of anything else, just work with that list. $\endgroup$ – lulu Jul 28 at 12:09
  • $\begingroup$ and you need to consider only those that are multiples of $10$ $\endgroup$ – J. W. Tanner Jul 28 at 12:23
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From the given,

$$a=10n,b=10m$$ where $n,m$ are relative primes and $$10nm=100.$$

Hence from the factorizations of $10$, the solutions

$$10,100;20,50;50,20;100,10.$$

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WLOG

$\dfrac aA=\dfrac bB=10;(A,B)=1$

$[a,b]=[10A,10B]=10[A,B]=100$

$\implies[A,B]=?$ with $(A,B)=1$

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