hcf of two number is12 and their lcm is 720.what may be possible two numbers
Answers
Step-by-step explanation:
The hcf and lcm of two numbers are 12 & 720, how many pairs are possible and what may be those?
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For any given values of hcf and lcm, here is a technique that will always get you a correct answer:
First, take the lcm and divide it by the hcf.
720÷12=60
Factor the number you got from that division into prime factors.
60=22⋅3⋅5
To find the number of possible pairs, count the number of primes, ignoring multiplicity. For this example, there are 3 primes: 2, 3, and 5.
The number of possible pairs will be 2 raised to the power of (the number of primes minus 1). So in this case, the number of possible pairs is: 23−1=4 .
To determine what those pairs are, start with the hcf and multiply it by one of your primes (include the multiplicity this time). For this example, I will start with the factor of 2, so 12⋅22=48 . Use that and the hcf to make a set of 2 numbers:
{48,12}
Now, take another one of the primes, and multiply one of the numbers in the set by it. There are 2 ways of doing this, so it will give you 2 sets of numbers, I will choose 3 as the next prime:
{{3⋅48,12},{48,3⋅12}}={{144,12},{48,36}}
From here, you repeat the process. For each set of 2 numbers, multiply one of them by one of the remaining primes, which means each set of 2 will be duplicated. The only prime left for me to deal with is 5 in this case:
{{5⋅144,12},{144,5⋅12},{5⋅48,36},{48,5⋅36}}={{720,12},{144,60},{240,36},{48,180}}
Therefore, for this problem, the possible pairs of numbers are 720 and 12, 240 and 36, 180 and 48, and 144 and 60.
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The problem is equivalent to ask: How many pairs of coprime numbers are possible (and what may be those) if their lcm is 60 ?
60=22⋅3⋅5 ,
and each of these three factors must be put as a whole into the two integers, n and m . Let’s consider ordered pairs (i.e., the pair n=1 , m=60 is considered different from the pair n=60 , m=1 ). Each of the three factors can be placed into 2 different positions ( n or m ) and therefore we have 23 possibilities, i.e., 8 different ordered pairs. If we consider only the numbers and not the order, the possibilities are 4 , as in the previous calculation every distribution is counted twice.
These 4 not-ordered pairs are
1, 60
4, 15
3, 20
5, 12
which correspond (in the original problem) to
12, 720
48, 180
36, 240
60, 144.
HCF*LCM=product of two numbers. Therefore
12*720= product of two numbers
Factors of HCF and LCM are
12=2*2*3 and
720= 12*5*3*2*2.
So two numbers can be
(12,720),(24,360),(36,240),
(48,180),(60,144),(72,120)
but in
(24,360),(72,120)
HCF will increase,
so rejecting these
required numbers are
(12,720), (36,240), (48,180), (60,144)
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The LCM and HCF of two numbers are 12 and 72, and the addition of two numbers is 60. What are the numbers?
let the numbers are x and y
also let x=hcf*a and y=hcf*b
then xy=hcf * lcm
(hcf)^2 *ab=hcf*lcm
hcf*ab=lcm
ab=lcm/hcf=720/12=60
ab=60
b=60/a
now as divisors of 60 are =2,3,4,5,6,10,12,15,20,30 ( 10 )
possible values of (a,b)
(2,30),(3,20),(4,15),(5,20),(6,10) 5 combinations
hence possible values of (x,y) are:
(24,360), ( 36,240), (48,180),(60,240),(72,120) ……..5 pairs
But here HCF(24,360)=24
Thus possible pairs are ( 36,240), (48,180),(60,240),(72,120) ……..4 pairs