Given that HCF (161,345) = 23, find their LCM.

Asked by Last Modified  

15 Answers

Learn Mathematics +1

Follow 17
Answer

Please enter your answer

CSAT Tutor, BPSC DI Tutor (Ex. Drishti IAS Senior Content Writer)

N¹ *N² = LCM * HCF 161 *345 =23 * N² N² =2415
Comments

Maths Tutor

We know that lcm x hcf = product of the numbers Therefore, lcm=161*345/23=161*15=2415
Comments

N1*N2=lcm*hcf (161*345)/23=2415 Lcm=2415
Comments

Anyone can learn the subject in easiest way, if they have good tutor to guide.

HCF*LCM = product of 2 numbers.So here 161*345 = 23*LCM, on solving we get LCM = 2415
Comments

LCM is 10.
Comments

Maths Expert

HCF*LCM= No. 1 *No. 2 Let the LCM be x 23*x = 161*345. On solving we get x= 2415
Comments

Tutor

LCM = (161*345)/23 = 2415.
Comments

COMPUTER SOLUTION

161*345/23
Comments

By using the formula HCF * LCM = Product of two numbers. Let the LCM be x 23 * x = 161 * 345 x = 161 * 345 / 23 = 2415 LCM of 161 and 345 is 2415
Comments

Formula to Note: 1) Product of two numbers = Product of their H.C.F. and L.C.M Solution: ⇒ Given that two numbers ,let a=161 and b=345; ⇒ HCF(161,345) = 23 ⇒ By applying the above formula,we get following expression ⇒ a * b = HCF*LCM ⇒ 161*345 = 23 * LCM(161,345) ⇒ 55545...
read more
Formula to Note: 1) Product of two numbers = Product of their H.C.F. and L.C.M Solution: ⇒Given that two numbers ,let a=161 and b=345; ⇒ HCF(161,345) = 23 ⇒ By applying the above formula,we get following expression ⇒ a * b = HCF*LCM ⇒ 161*345 = 23 * LCM(161,345) ⇒ 55545 = 23 * LCM(161,345) ⇒ Rewriting the above the expression ⇒ 23 * LCM(161,345) = 55545 ⇒ LCM(161,345) = 55545/23 ⇒ LCM(161,345) = 2415 ⇒ There for LCM of 161, 345 is 2415. read less
Comments

View 13 more Answers

Related Questions

Show that x = 2 and y = 1 satisfy the linear equation 2x 3y = 7.
2x + 3y=7 Put 2 in place of x and 1 in place of y. 2*2+ 3*1=4+3=7
Sajjan
If the radius of a sphere is 2 r, then what is its volume?
Volume of the sphere =( 4/3 π r3 ) cubic unit, If r = 2r then, Volume of the sphere: = = = 32/3 π r3.
Deva
Solve : 141x + 103y = 217; 103x + 141y = 27
First add both equations, we'll get 244x+244y=244, Now, Divide this equation by 244, it gives x+y=1.....(a) Again, subtract equation 2 from 1, it gives 38x-38y=190, Now, divide this equation by 38,comes x-y=5.....(b) Now...
Tanvi

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a Question

Related Lessons


MAXIMA AND MINIMA
QUESTION In a college, where every student follows at least one of the three activities- dance, music, or sports- 65% follow dance, 86% follow music, and 57% follow sports. What can be the maximum...



Recommended Articles

Quality education does not only help children to get a successful career and life, but it also hugely contributes to society. The formal education of every child starts from school. Although there are numerous schools, parents find it challenging to choose the right one that would fit their child. It is difficult for them...

Read full article >

Once over with the tenth board exams, a heavy percentage of students remain confused between the three academic streams they have to choose from - science, arts or commerce. Some are confident enough to take a call on this much in advance. But there is no worry if as a student you take time to make choice between - science,...

Read full article >

With the mushrooming of international and private schools, it may seem that the education system of India is healthy. In reality, only 29% of children are sent to the private schools, while the remaining head for government or state funded education. So, to check the reality of Indian education system it is better to look...

Read full article >

Learning for every child starts from a very young age. While the formal methods include school curriculums and private lessons, the informal methods include dancing, music, drawing, and various fun-filling activities. Playing games and practising these types of activities helps the children get out of boredom and...

Read full article >

Looking for Class 9 Tuition ?

Learn from the Best Tutors on UrbanPro

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you