The answer is 8.
24
103
62 is the smallest number that leaves a remainder of 2 when divided by 3, 4, 5, and 6.
Six.
0.4286
23
It is 30.
24
12
0.2308
The smallest whole number that can be divided by both 3 and 6 and leaves a remainder of 1 is 7.
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.