This is a LCM question, so factor all and find the highest power of each prime factor.
2 is prime
3 is prime
5 is prime
6 is [2 3]
7 is prime.
LCM is [2] [3] [5] [7] so the answer is 210◄
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Seven.
59
29
5. To leave a remainder of 1 when divided by 2, the number must be odd. To leave a remainder of 2 when divided by 3, the number must also be two more than a multiple of 3. The multiples of 3 which are odd are 1 x 3, 3 x 3, 5 x 3, etc. The smallest odd multiple of three is 1 x 3 = 3 ⇒ required number is 3 + 2 = 5.