The integer is 157.
157/3 = 52 remainder 1
157/5 = 31 remainder 2
157/7 = 22 remainder 3
forty-one
36
It is LCM(5, 8) + 1 = 41.
HCF(5, 8) + 1 = 40 + 1 = 41
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.
234568
5
41
41
41
121.
forty-one
36
71
It is LCM(5, 8) + 1 = 41.
HCF(5, 8) + 1 = 40 + 1 = 41
127 is the least prime number greater than 25 that will have a remainder of 2 when divided by 25.