3) Which two consecutive even
numbers have an LCM of 180?
O 15 and 13
O 12 and 20
18 and 20
O 14 and 28
Answers
Answer:
Given: Two consecutive even terms and LCM = 180
Let’s assume the two terms:
1st term : 2x
2nd term : 2x + 2
There is a rule which states that
If ‘a’ and ‘b’ are two numbers, then their product is equal to the product of their LCM and HCF.
i.e,
a∗b=HCF∗LCM - (i)
We can logically say that the HCF of any two consecutive even numbers is always equal to ‘2’ provided that the smallest no. is at least ‘2’.
i.e.,
HCF of ‘2x’ and ‘2x + 2’ is ‘2’ - (ii)
Therefore,
from (i) and (ii),
2x∗(2x+2)= HCF * LCM
2x∗(2x+2)=2∗180
x∗(2x+2)=180
2x2+2x−180=0
Solving the quadratic equation we get,
x=9,−10 (we discard -10 since it is negative)
Therefore, the numbers are
2∗9=18 (1st no.)
2∗9+2=18+2=20 (2nd no.)
Answer : 18,20
Thank You,
Since, the required terms are consecutive even numbers, we can represent them as k= 2x,l=2(x+1).
We know that, the product of two numbers is equal to the product of their HCF and LCM.
Therefore,
k∗l=HCF∗LCM
=>2x∗2(x+1)=HCF∗LCM
But, it is clear that the HCF of two consecuive even integers is 2.
=> 2x(2x+2)=2∗LCM
But, it is given that LCM is 180
=>2x2+2x=180
=>x2+x−90=0
=>x2+10x−9x−90=0
=>x(x+10)−9(x+10)=0
Therefore, x is either 9or−10
If x is negative, k and l would be negative and the notion of LCM would make no sense. So, we take x to be 9 only.
Substituting the value of x in k and l
k=2x=2(9)=18
l=2(x+1)=2(9+1)=2(10)=20
So the required numbers are 18 and 20
Let the two consecutive even numbers are 2n and 2n + 2. We are to get LCM of 2n & 2(n+1).
2n = 2*n
2(n+1) = 2*(n+1) . Hence their LCM is = 2*n*(n+1) According to condition,
2n(n+1) = 180, dividing both side by 2
n(n+1) =90
or, n^2 + n -90 =0
or, n^2+ 10n -9n -90 =0
Or, n(n +10) -9(n+10)=0
or, (n+10)(n-9) =0, Therefore
Either n+10 =0 , or n- 9=0 ,
So, n= - 10 or 9 , As the consecutive even numbers are tske.Therefore consecutive even numbers are taken 2n and 2n +2 ,they are , {2*(-10) , 2*(-10 ) +2}, i.e. (-20, -18)
or ,{ 2*9 , 2*9 +2} i.e. (18, 20)