Q: Is 1242 divisible by 23

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2 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.

Related questions

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

All 16 of these: 1 2 3 6 9 18 23 27 46 54 69 138 207 414 621 1242.

No. 23 is not divisible by anything except itself and one.

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.