We are looking for the lowest common multiple of 3, 4 & 5. This is 60 (3 * 20, 4 * 15 & 5 * 12). We add 2 to this so that the remainder will be 2 on division.
Therefore the smallest integer that meets these requirements is 62.
71
127 is the least prime number greater than 25 that will have a remainder of 2 when divided by 25.
The smallest prime number greater than 200 is 211.
1009 is the smallest prime number greater than 1000.
what is the smallest prime number that is greater than 50
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5
41
The integer is 157. 157/3 = 52 remainder 1 157/5 = 31 remainder 2 157/7 = 22 remainder 3
121.
The remainder is the number that is left over after the initial value has been divided as much as it can. If any numbers greater than 48 were present as a remainder, then these could be divided further into 48. If 48 is present as the remainder, then this can be divided by 48 to give 1, leaving no remainder. Thus, the largest possible remainder if the divisor is 48 is 47.
36
If it is divided by a fraction or a decimal. Like 1/5 or .986
71
127 is the least prime number greater than 25 that will have a remainder of 2 when divided by 25.
Absolutely not possible
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