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# lcm of 20 and 38

hfc of 20 and 38 lcm of 20 and 38

## Research, Knowledge and Information :

### What is the least common multiple of 20 and 38 - Answers

What is the least common multiple of 20 and 38? ... The least common multiple of 38 and 57 is 114. 1 person found this useful Eric Barnes. 390,708 Contributions.

### Least Common Multiple of 20 and 38 - Times Table

The lcm of 20 and 38 is the smallest positive integer that divides the numbers 20 and 38 without a remainder. Spelled out, it is the least common multiple of 20 and 38.

### Least Common Multiple (LCM) of 20 and 38 - Equations solver

Least Common Multiple (LCM) of 20 and 38 . If it's not ... To find the least common multiple of two numbers just type them in and get the solution.

### Least Common Multiple of 20 and 38 LCM(20,38) - gcflcm.com

... find the lowest common denominator (lcd) smallest multiple of two integers, calculate prime factors and least common multiple of 20 and 38. Home;

### LCM of 20 and 38 - coolconversion.com

See how to find the Least Common Multiple of any number using our Least Common Multiplier (LCM ... Here is the answer to questions like: LCM of 20 and 38 or What is ...

### What is the LCM of 38 and 20 - Answers.com

What is the LCM of 20 and 30 and 40? The LCM is: 120 ... Multipying the common fctor back, the LCM of 38 and 57 is 6*19 = 114. 1 person found this useful

### Greatest Common Factor of 20 and 38 - GCF and LCM Calculator

Greatest Common Factor of 20 and 38. Greatest common factor (GCF) ... Also check out the Least Common Multiple of 20 and 38. Related Greatest Common Factors of 20.

### Least Common Multiple of 20 36 38 and 41

To find the Least Common Multiple or LCM of 20, 36, 38 and 41, ...

### Least Common Multiple (LCM) of 28 and 38 - Equations solver

Least Common Multiple (LCM) of 28 and 38 . If it's not ... Least Common Multiple (LCM) of 20 and 13 | | Least Common Multiple (LCM) of 42 and 3 | ...

### Least common multiple - Wikipedia

... 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64 ... A least common multiple of a and b is a common multiple that is minimal in the sense that for any other ...

## Suggested Questions And Answer :

### Explain when the LCM of 2 expressions is equal to all of both of them multiplied together.

It just means that there are no common factors. For example, the LCM of 5 and 7 is 5*7=35 because 5 and 7 have no common factors. But 10 and 14 do have a common factor 2, so the LCM is 2*5*7=70 not 140=10*14. You could say that the LCM of a and b is ab/HCF(a,b). In other words a=hm and b=hn where h is the HCF and m and n are integers. LCM=hm*hn/h=mnh. If a is a multiple of b then we can write LCM=ab/b=a. To work out the LCM of more than two integers, take a pair and calculate the LCM then take the next integer and calculate the LCM of this integer with the LCM of the first two integers, and so on. So we get LCM(a,b,c)=LCM(LCM(a,b),c), LCM(LCM(a,c),b) or LCM(LCM(b,c),a). For example: 4, 18, 12: LCM(4,18)=4*18/2=36; LCM(36,12)=36*12/12=36. So the LCM of 4, 18 and 12 is 36. Or LCM(18,12)=18*12/6=36; LCM(36,4)=36*4/4=36. Or LCM(4,12)=4*12/4=12; LCM(12,18)=12*18/6=36.

### whats the lcm of 4,9,13?

When dealing with several numbers, the lowest(or least) common multiple can be applied consecutively, i.e. lcm( 4,9,13) =lcm(4, lcm(9,13)) (or lcm(lcm(4,9), 13) (i.e. find the lowest common  of any 2 numbers, and then find the lowest common multiple of the result and the remaining number) lcm(9,13) = 9 x 13 = 117 (13 is a prime, so no chance of common factors) http://statsfiddle.info/Primes/LowestCommonMultiple/9?number2=13 lcm(117,4) = 468 Again, there are no common factors http://statsfiddle.info/Primes/LowestCommonMultiple/117?number2=4 i.e. lcm(4, 9,13) = 468 In your example, all 3 numbers are co-prime, meaning there are no common factors at all.

### Find the LCM 21c^3b^8,7c^2b^3,15c^5b^7

first, the numbers the lcm of 21, 7, and 15 21 = 3 * 7 7 = 7 15 = 3 * 5 for an lcm you times together a set of numbers that includes enough numbers to make each of those factors (???) like this: to make 21 you need a 3 and a 7 to make 7 you just need 7 to make 15 you need a 3 and a 5 to make the lcm, you want 3 * 5 * 7 because 3 * 5 * 7 has a 3 * 7 (to make 21), a 7 (to make 7), and a 3 * 5 (to make 15). the lcm of 21, 7, and 15 is 3 * 5 * 7 = 105 now the c's the lcm of c^3, c^2, and c^5 is c^5 now the b's the lcm of b^8, b^3, and b^7 is b^8 the lcm of the whole thing is 105(c^5)(b^8)

### 15a^3b^4, 42a^2b^5c (find LCM)

first, the numbers: the lcm of 15 and 42 is 15 = 3 * 5 42 = 2 * 3 * 7 we want a 2 we want one 3  (because they both have it already, we don't need two 3's) we want a 5 and a 7 2 * 3 * 5 * 7 = 210 the lcm of 15 and 42 is 210 now the a's the lcm of a^3 and a^2 is a^3 note:  a^3 = a^2 * a now the b's the lcm of b^4 and b^5 is b^5 now the c's the lcm of 1 and c is c note:  since 15(a^3)(b^4) doesn't have a c, it's like there's a *1 sitting there the lcm of the whole thing is 210(a^3)(b^5)c

### What's the LCM of 20x cubed and 16x to the 4 power?

if you mean 20x^3 and 16x^4 then first, the numbers 20 = 2 * 2 * 5 16 = 2 * 2 * 2 * 2 the lcm is 2 * 2 * 2 * 2 * 5 because 2 * 2 * 2 * 2 * 5 contains 2 * 2 * 5 (to make 20) and 2 * 2 * 2 * 2 (to make 16). 2 * 2 * 2 * 2 * 5 = 80 the lcm of 20 and 16 is 80 now the x's the lcm of x^3 and x^4 is x^4 the lcm of the whole thing is 80x^4   if you mean (20x)^3 and (16x)^4 then: (20x)^3 = 8000x^3 (16x)^4 = 65536x^4 first, the numbers 8000 = 2^6 * 5^3 65536 = 2^16 the lcm is 2^16 * 5^3 because 2^16 * 5^3 contains 2^6 * 5^3 (to make 8000) and 2^16 (to make 65536) 2^16 * 5^3 = 8192000 now the x's the lcm of x^4 and x^3 is x^4 the lcm of the whole thing is 8192000x^4

### what is the LCM and the GCF of 125, 235 and 270

LCM of 125, 235, 270 Factor each into primes: 125 = 5 * 5 * 5 235 = 5 * 47 270 = 2 * 3 * 3 * 3 * 5 The LCM is the product of the minimum list of primes needed to make all of those numbers (125, 235, 270). For instance, there's a factor of 2 in 270, so the LCM has to have a factor of 2. There are three 3's as factors in 270, so the LCM has to have a factor of three 3's. There are three 5's as factors in 125, so the LCM has to have a factor of three 5's.  There's a 5 in 235, but that's already covered by one of the three 5's. There's a 47 in 235, so the LCM has to have a factor of 47. Basically you start with this: 125 = 5 * 5 * 5 235 = 5 * 47 270 = 2 * 3 * 3 * 3 * 5 Then make a list like this: 2 * 3 * 3 * 3 * 5 * 5 * 5 * 27 Then multiply it together like this: 2 * 3 * 3 * 3 * 5 * 5 * 5 * 27 = 182250 Answer:  The LCM of 125, 235, and 270 is 182250 . GCF of 125, 235, and 270 Start with a list of primes: 125 = 5 * 5 * 5 235 = 5 * 47 270 = 2 * 3 * 3 * 3 * 5 The GCF is the product of the list of primes that appear in all of those numbers (125, 235, 270). 47 only appears as a factor in 235, so 47 does not appear in the GCF. The only number that appears in all three numbers (125, 235, 270) is 5. 5 appears in 125 three times, but it only appears once in all three numbers (125, 235, 270), so the GCF only includes one 5. Answer:  The GCF of 125, 235, and 270 is 5.

### whats the lcm for 2a and 2b

LCM="leest kommon multiplier" lcm(2a,2b) depend on a & b if smaller av (2a,2b) divide intu other with leftover=0, lcm=smaller zampel...a=2 & b=4, then hav 4 & 8, but 4 go intu 8 with no leftover, so LCM=4 but if 2a=3 & 2b=7, LCM=21=3*7=4ab

### some exercise books have to be arranged in piles with 10or15 or 20 book s per pile what is the least number of books needed to do this?

To make one pile you would only need 10 books. The question doesn't say how many piles. It only says that you can't have more than 20. It also says or appears to say that you can't have a pile with less than 10 books. And "or" means you have a choice. So if, for example, you had 15 books you couldn't make a pile of 10 and a pile of 5, you would have to have a pile of 15; but if you had 20 books you could have two piles of 10 or one pile of 20. It looks like the minimum would be 10 because you couldn't have less than 10 to make a pile. Now, let's look at another interpretation of the question. The LCM of 10, 15, 20 (5*2, 5*3, 5*4) is the LCM of 2, 3, 4 multiplied by the HCF 5. The LCM of 2, 3, 4 is 12 so the LCM of 10, 15, 20 is 5*12=60. 60 books can be piled in different ways: 6 piles of 10, 4 piles of 15, or 3 piles of 20. Or they could be piled as 2 piles of 10 and 2 piles of 20; or 3 piles of 10 and 2 piles of 15, etc. The question doesn't say if the piles have to be all the same size. The minimum number of books that can be piled in piles of 5, 15 or 20 books a pile is therefore 60. So it depends on how you interpret the question. If you have been working with HCF and LCM recently it's likely that the question has been given to you to apply your learning to a problem. So the context of the question may be significant and may guide you to the intended interpretation. If you have been learning about LCM and HCF then go for 60 as the answer.

### What is the LCM of (x+4)(x+2)and(x-2)(x+2)

What is the LCM of (x+4)(x+2)and(x-2)(x+2) Let M = (x+4)(x+2), and let N = (x-2)(x+2) Take out the common factor of (x+2) This gives us, M= F*(x+4), and N = F*(x-2) Then lcm(N,N) = F*lcm((x+4), (x-2)) = F*{(x+4)*(x-2)} lcm = (x+2)(x+4)(x-2) = (x+4)(x^2 - 4) Answer: lcm = (x+4)(x^2 - 4), where (x+4) and (x-2) are coprime.

### LCM of 15 c^2b^6, 5c^6b^4, 6c^7b^3

numbers first: lcm of 15, 5, and 6 is 30 next the c's lcm of c^2, c^6, and c^7 is c^7 note: c^7 = c^2 * c^5 and c^7 = c^6 * c next the b's the lcm of b^6, b^4, and b^3 is b^6 the lcm for the whole thing is 30(c^7)(b^6) note: you don't have to write it with parentheses.  there's no easy way to do exponents on this page, so I wrote it with parentheses here.