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What is 2484 evenly divisible by?
2 and any multiple thereof up to 1242.
What is the prime factorization of 1242?
The PF of 1242 is: 2 x 3 x 3 x 3 x 23
Is 23 divisible by3?
23 is a prime number so it is not divisible by 3. It is only divisible by 1 and 23. Dividing 23 by three gives the result of 7.6r
What number is not divisible 7?
23 is not divisible by 7.
What is the least number which divides 285 and 1249 leaving remainders 9 and 7?
To solve this you need the first common factor of 285 - 9 = 276 and 1249 - 7 = 1242 which is greater than 9 (if it was 9 or less it could NOT leave a remainder of 9 when divided into 285): Common factors of 276 and 1242 are the factors of their hcf: 276 = 2² x 3 x 23 1242 = 2 x 3² x 23 hcf = 2 x 3 x 23 = 138 The factors of 138 which are the common factors of 276 and 1242 are: 1, 2, 3, 6, 23, 46, 69, 138 The first one greater than 9 is 23 →23 is the least number which divides 285 and 1249 to leave remainders of 9 and 7 respectively.
What is 2484 evenly divisible by?
2 and any multiple thereof up to 1242.
What is the least common multiple of 23 and 54?
The LCM of 23 and 54 is 1242.
What is the prime factorization of 1242?
The PF of 1242 is: 2 x 3 x 3 x 3 x 23
What are the factors of 1242?
1, 2, 3, 6, 9, 18, 23, 27, 46, 54, 69, 138, 207, 414, 621, 1242
What numbers is 23 divisible by?
Only 1 and 23.
What is the prime factor tree of 1242?
Factorization: 2*3*3*3*23
What are the prime factors of 1242?
(including each only once) 2, 3, 23
Is 23 divisible by3?
23 is a prime number so it is not divisible by 3. It is only divisible by 1 and 23. Dividing 23 by three gives the result of 7.6r
What numbers can divide 1242 evenly?
All 16 of these: 1 2 3 6 9 18 23 27 46 54 69 138 207 414 621 1242.
Is 23 divisible by 5?
No. 23 is not divisible by anything except itself and one.
What number is not divisible 7?
23 is not divisible by 7.
What is the least number which divides 285 and 1249 leaving remainders 9 and 7?
To solve this you need the first common factor of 285 - 9 = 276 and 1249 - 7 = 1242 which is greater than 9 (if it was 9 or less it could NOT leave a remainder of 9 when divided into 285): Common factors of 276 and 1242 are the factors of their hcf: 276 = 2² x 3 x 23 1242 = 2 x 3² x 23 hcf = 2 x 3 x 23 = 138 The factors of 138 which are the common factors of 276 and 1242 are: 1, 2, 3, 6, 23, 46, 69, 138 The first one greater than 9 is 23 →23 is the least number which divides 285 and 1249 to leave remainders of 9 and 7 respectively.
