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Q: What s the LCM of 2 3 and 4?

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4=2*2 12=2*2*3 14=2*7 4 goes into 12, so the lcm of 4,12, and 14 is equal to the lcm of 12 and 14 12=2*2*3 14=2*7 We can cancel the 2's and just have 2*2*3*7=84 84.

15 = 3 x 5 6 = 2 x 3 25 = 5 x 5 Therefore, the LCM is the product of a 2, a 3, and two 5's. 2 x 3 x 5 x 5 = 150

Start by factoring both numbers into primes: 210 = 2*3*5*7 450= 2*3*3*5*5 What's the shortest list that includes all the factors of both numbers? One 2, two 3's, two 5's, one 7 That's the recipe for lcm in this problem. lcm = 2*3*3*5*5*7 = 3150

36

4725.189 = 3*63 = 3*3*21 = 3*3*3*7.75 = 3*25 = 3*5*5.The LCM needs 3 3's, 2 5's and a 7. ... (Leave out the 3 from 75 since there are already 3's in 189.)LCM = 3*3*3*5*5*7 = 27*25*7 = 675*7 = 4725.

240 = 2 x 2 x 2 x 2 x 3 x 5 900 = 2 x 2 x 3 x 3 x 5 x 5 Max is four 2's, two 3's and two 5's So LCM is 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 3600

The LCM of 4, 8, and 56 is 6. Since 4 and 8 are factors of 56, 56 is the LCM of the set. The answer can be found by dividing the 3 numbers into their prime factors, in this case 4=2x2, 8=2x2x2, and 56=7x2x2x2. The lowest common multiple is the smallest number which has all of these prime factors in. That means it has to have two 2's in (from 4) three 2's in (from 8) and three 2's and a 7 from 56. In this case, all of the prime factors of 4 can be found in 8, and all of the prime factors of 8 are in 56, so the LCM (lowest common multiple) of these 3 is 56.

The LCM of 4 and 15 is 60.The least common multiple is the smallest number that is multiple of two or more numbers.Multiples of 4: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60Multiples of 15: 15, 30, 45, 60The LCM of 4 and 15 is 60.The LCM must include the highest amount of each prime factor in both numbers. The highest power of 2's is 2, of 3's is 1 and of 5's is 1.So the LCM = 2 x 2 x 3 x 5 = 6060 15x1=15 15x2=30 15x3=45 15x4=60

It is: 140

2/3 = 14/21 4/7 = 12/21 So 2/3 IS bigger.

2 3/4= 2.75

well do the math: 1's 1*4(# of suits)=4 2's 1*4(# of suits)=4 3's 1*4(# of suits)=4 add them all up and you get 12 so you have 12 1's 2's and 3's in a deck of 52 playing cards.

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