It is LCM(3, 4, 5, 6) + 2 = 60 + 2 = 62
The smallest number that leaves a remainder of 2 when divided by 3, 4, 5 or 6 is 62. This can be found by analyzing the multiples of the greatest common divisor (GCD) of 3, 4, 5, and 6, which is 60. Adding 2 to this number gives us the smallest number with the specified remainder.
The integer is 157. 157/3 = 52 remainder 1 157/5 = 31 remainder 2 157/7 = 22 remainder 3
23
6172839
Try 30
85
234568
41
121.
62 is the smallest number that leaves a remainder of 2 when divided by 3, 4, 5, and 6.
The integer is 157. 157/3 = 52 remainder 1 157/5 = 31 remainder 2 157/7 = 22 remainder 3
0.25
23
6172839
Try 30
36 divided by 28 would be 1 remainder 8and44 divided by 32 would be 1 remainder 12
85
12*1 + 3 = 15 12*2 + 3 = 27 12*4 + 3 = 51 when 51 divided by 15 gives the remainder as 6.