Yes - 1242/23 = 54
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∙ 12y ago2 and any multiple thereof up to 1242.
The PF of 1242 is: 2 x 3 x 3 x 3 x 23
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
23 is not divisible by 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.
2 and any multiple thereof up to 1242.
The LCM of 23 and 54 is 1242.
The PF of 1242 is: 2 x 3 x 3 x 3 x 23
1, 2, 3, 6, 9, 18, 23, 27, 46, 54, 69, 138, 207, 414, 621, 1242
Only 1 and 23.
(including each only once) 2, 3, 23
Factorization: 2*3*3*3*23
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
No. 23 is not divisible by anything except itself and one.
All 16 of these: 1 2 3 6 9 18 23 27 46 54 69 138 207 414 621 1242.
23 is not divisible by 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.