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The $HCF$ and $LCM$ of two numbers are $12$ and $720$. lets try to find how many pairs are possible and what may be those number.

Cant get to the answer given help please

$Answer :$ $4$ $pairs$ $(12,720) , (36,240) , (48,180) ,(60,144).$

Pallab
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  • One useful fact is that the product of the HCF (also called greatest common divisor) and the LCM is the product of the two numbers. – hardmath Mar 10 '17 at 16:41

3 Answers3

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Write $12=2^2\times 3$ and $720=2^4\times 3^2\times 5$. We must have $x=2^a3^b5^c$ and $y=2^d3^e5^f$, since they can't be divisible by any other primes ($a,b,c,d,e,f\geq 0$). Now to get the lcm of two numbers you take the larger power of each prime, and to get the hcf you take the smaller power. So $\max(a,d)=4$ and $\min(a,d)=2$, so $a$ and $d$ are $2$ and $4$ in some order. Similar reasoning applies to $b,e$ and $c,f$. Each of these three pairs may be swapped independently.

Especially Lime
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Call the numbers $a$ and $b$, and $a'=a/12, b'=b/12$. Then $a' b' = 720/12 = 60$. Then $60=4\cdot 3\cdot 5$, and we need to split those three factors (we cannot split the $4$) between $a'$ and $b'$, so there are $8$ ordered options, $4$ unordered options: $(1,60), (4,15), (3,20), (5,12)$.

Then multiply through by $12$ to recover $(a,b)$.

Joffan
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$A = a*12$ and $B = a*12$ and $a$ and $b$ are relatively prime.

LCM = $720 = ab*12$ so $ab = 60=2^2*3*5$. So $a = 2^c3^d5^e$ and $b= 2^{2-c}3^{1-d}5^{e-1}$ where $c = 0$ or $2$. $d= 0$ or $1$. $e=0$ or $1$. That is $2^3=8$ choices. As we don't want two pairs of the same numbers in different orders, there are $8/2 =4$ pairs. They are:

$a = 0;b=0; c=0: a=1;b= 60: A =12; B=720$

$a=0;b=0;c=1; a=5;b=12: A=60; B=144$

$a=0;b=1;c=0; a=3; b=20: A=36; B=240$

$a=0;b=1;c=1; a=15; b=4: A=180; B=48$

fleablood
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