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How do you find factors of 3150? - Answers
There are a few ways to go about factoring. You can decide what works best for you. I always find the prime factorization first. Let's look at a random number: 3150 The prime factorization can be found by using a factor tree. 3150 1575,2 525,3,2 175,3,3,2 35,5,3,3,2 7,5,5,3,3,2 2 x 3^2 x 5^2 x 7 = 3150 Half of the factors will be less than the square root, half greater. If the number is a perfect square, there will be an equal number of factors on either side of the square root. In this case, the square root is between 56 and 57. Adding one to the exponents of the prime factorization and multiplying them will tell you how many factors there are. In this case, the exponents are 1, 2, 2 and 1. Add one to each. 2 x 3 x 3 x 2 = 36 3150 has 36 factors.18 of them are 56 or less, 18 of them are 57 or greater. All we have to do is divide the numbers 1 through 56 into 108. If the result (quotient) turns out to be an integer, you've found a factor pair. Knowing the rules of divisibility will make that even easier. 3150 is divisible by... 1 because everything is. 2 because it's even. 3 because its digits add up to a multiple of 3. 5 because it ends in zero. 6 because it's a multiple of 2 and 3. 7 because it's in the prime factorization. 9 because its digits add up to a multiple of 9. 10 because it ends in zero. That's eight. We need ten more. They can be found in the prime factorization. 2 x 7 = 14 3 x 5 = 15 3^2 x 2 = 18 3 x 7 = 21 5^2 = 25 2 x 3 x 5 = 30 5 x 7 = 35 2 x 3 x 7 = 42 3^2 x 5 = 45 2 x 5^2 = 50 That's 18 factors less than 56. Divide them into 3150. That's the rest of them. (3150, 1) (1575, 2) (1050, 3) (630, 5) (525, 6) (450, 7) (350, 9) (315, 10) (225, 14) (210, 15) (175, 18) (150, 21) (126, 25) (105, 30) (90, 35) (75, 42) (70, 45) (63, 50) 1, 2, 3, 5, 6, 7, 9, 10, 14, 15, 18, 21, 25, 30, 35, 42, 45, 50, 63, 70, 75, 90, 105, 126, 150, 175, 210, 225, 315, 350, 450, 525, 630, 1050, 1575, 3150 Notice that all of those numbers, except for 1, can also be found in the prime factorization.
Bing
How do you find factors of 3150? - Answers
There are a few ways to go about factoring. You can decide what works best for you. I always find the prime factorization first. Let's look at a random number: 3150 The prime factorization can be found by using a factor tree. 3150 1575,2 525,3,2 175,3,3,2 35,5,3,3,2 7,5,5,3,3,2 2 x 3^2 x 5^2 x 7 = 3150 Half of the factors will be less than the square root, half greater. If the number is a perfect square, there will be an equal number of factors on either side of the square root. In this case, the square root is between 56 and 57. Adding one to the exponents of the prime factorization and multiplying them will tell you how many factors there are. In this case, the exponents are 1, 2, 2 and 1. Add one to each. 2 x 3 x 3 x 2 = 36 3150 has 36 factors.18 of them are 56 or less, 18 of them are 57 or greater. All we have to do is divide the numbers 1 through 56 into 108. If the result (quotient) turns out to be an integer, you've found a factor pair. Knowing the rules of divisibility will make that even easier. 3150 is divisible by... 1 because everything is. 2 because it's even. 3 because its digits add up to a multiple of 3. 5 because it ends in zero. 6 because it's a multiple of 2 and 3. 7 because it's in the prime factorization. 9 because its digits add up to a multiple of 9. 10 because it ends in zero. That's eight. We need ten more. They can be found in the prime factorization. 2 x 7 = 14 3 x 5 = 15 3^2 x 2 = 18 3 x 7 = 21 5^2 = 25 2 x 3 x 5 = 30 5 x 7 = 35 2 x 3 x 7 = 42 3^2 x 5 = 45 2 x 5^2 = 50 That's 18 factors less than 56. Divide them into 3150. That's the rest of them. (3150, 1) (1575, 2) (1050, 3) (630, 5) (525, 6) (450, 7) (350, 9) (315, 10) (225, 14) (210, 15) (175, 18) (150, 21) (126, 25) (105, 30) (90, 35) (75, 42) (70, 45) (63, 50) 1, 2, 3, 5, 6, 7, 9, 10, 14, 15, 18, 21, 25, 30, 35, 42, 45, 50, 63, 70, 75, 90, 105, 126, 150, 175, 210, 225, 315, 350, 450, 525, 630, 1050, 1575, 3150 Notice that all of those numbers, except for 1, can also be found in the prime factorization.
DuckDuckGo
How do you find factors of 3150? - Answers
There are a few ways to go about factoring. You can decide what works best for you. I always find the prime factorization first. Let's look at a random number: 3150 The prime factorization can be found by using a factor tree. 3150 1575,2 525,3,2 175,3,3,2 35,5,3,3,2 7,5,5,3,3,2 2 x 3^2 x 5^2 x 7 = 3150 Half of the factors will be less than the square root, half greater. If the number is a perfect square, there will be an equal number of factors on either side of the square root. In this case, the square root is between 56 and 57. Adding one to the exponents of the prime factorization and multiplying them will tell you how many factors there are. In this case, the exponents are 1, 2, 2 and 1. Add one to each. 2 x 3 x 3 x 2 = 36 3150 has 36 factors.18 of them are 56 or less, 18 of them are 57 or greater. All we have to do is divide the numbers 1 through 56 into 108. If the result (quotient) turns out to be an integer, you've found a factor pair. Knowing the rules of divisibility will make that even easier. 3150 is divisible by... 1 because everything is. 2 because it's even. 3 because its digits add up to a multiple of 3. 5 because it ends in zero. 6 because it's a multiple of 2 and 3. 7 because it's in the prime factorization. 9 because its digits add up to a multiple of 9. 10 because it ends in zero. That's eight. We need ten more. They can be found in the prime factorization. 2 x 7 = 14 3 x 5 = 15 3^2 x 2 = 18 3 x 7 = 21 5^2 = 25 2 x 3 x 5 = 30 5 x 7 = 35 2 x 3 x 7 = 42 3^2 x 5 = 45 2 x 5^2 = 50 That's 18 factors less than 56. Divide them into 3150. That's the rest of them. (3150, 1) (1575, 2) (1050, 3) (630, 5) (525, 6) (450, 7) (350, 9) (315, 10) (225, 14) (210, 15) (175, 18) (150, 21) (126, 25) (105, 30) (90, 35) (75, 42) (70, 45) (63, 50) 1, 2, 3, 5, 6, 7, 9, 10, 14, 15, 18, 21, 25, 30, 35, 42, 45, 50, 63, 70, 75, 90, 105, 126, 150, 175, 210, 225, 315, 350, 450, 525, 630, 1050, 1575, 3150 Notice that all of those numbers, except for 1, can also be found in the prime factorization.
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- og:descriptionThere are a few ways to go about factoring. You can decide what works best for you. I always find the prime factorization first. Let's look at a random number: 3150 The prime factorization can be found by using a factor tree. 3150 1575,2 525,3,2 175,3,3,2 35,5,3,3,2 7,5,5,3,3,2 2 x 3^2 x 5^2 x 7 = 3150 Half of the factors will be less than the square root, half greater. If the number is a perfect square, there will be an equal number of factors on either side of the square root. In this case, the square root is between 56 and 57. Adding one to the exponents of the prime factorization and multiplying them will tell you how many factors there are. In this case, the exponents are 1, 2, 2 and 1. Add one to each. 2 x 3 x 3 x 2 = 36 3150 has 36 factors.18 of them are 56 or less, 18 of them are 57 or greater. All we have to do is divide the numbers 1 through 56 into 108. If the result (quotient) turns out to be an integer, you've found a factor pair. Knowing the rules of divisibility will make that even easier. 3150 is divisible by... 1 because everything is. 2 because it's even. 3 because its digits add up to a multiple of 3. 5 because it ends in zero. 6 because it's a multiple of 2 and 3. 7 because it's in the prime factorization. 9 because its digits add up to a multiple of 9. 10 because it ends in zero. That's eight. We need ten more. They can be found in the prime factorization. 2 x 7 = 14 3 x 5 = 15 3^2 x 2 = 18 3 x 7 = 21 5^2 = 25 2 x 3 x 5 = 30 5 x 7 = 35 2 x 3 x 7 = 42 3^2 x 5 = 45 2 x 5^2 = 50 That's 18 factors less than 56. Divide them into 3150. That's the rest of them. (3150, 1) (1575, 2) (1050, 3) (630, 5) (525, 6) (450, 7) (350, 9) (315, 10) (225, 14) (210, 15) (175, 18) (150, 21) (126, 25) (105, 30) (90, 35) (75, 42) (70, 45) (63, 50) 1, 2, 3, 5, 6, 7, 9, 10, 14, 15, 18, 21, 25, 30, 35, 42, 45, 50, 63, 70, 75, 90, 105, 126, 150, 175, 210, 225, 315, 350, 450, 525, 630, 1050, 1575, 3150 Notice that all of those numbers, except for 1, can also be found in the prime factorization.
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