0, 1, 2 and 4.
0, 1, 2 and 4.
0, 1, 2 and 4.
0, 1, 2 and 4.
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The cannot be such a number. The biggest possible remainder is 2.
x = 2 mod 3; x = 3 mod 5. Equivalently, x = 3m + 2 = 5n + 3. One such number is 23; others could be found.
If the primes are 5 or greater, then the remainders are 1 or 5.This is so trivially obvious.The remainder cannot be 0 or else the number is divisible by 6 and so not a prime.The remainder cannot be 2 or 4 or else the number is divisible by 2 and so not a prime.The remainder cannot be 3 or else the number is divisible by 3 and so not a prime.That just leaves 1 and 5: 11 leaves a remainder of 5, 13 leaves 1 for example.
For a number to be divisible by another number it must be at least that number. As 3 is less than 605 it cannot be divisible by 605. Also 3 is not a FACTOR of 605, that is 3 does not divide into 605 without remainder.
Since the remainder is 0 when the numbers are divided by 3, then that number is a multiples of 3. For example, 45/3 = 15 remainder 0 45/4= 11 remainder 1 45/7 = 6 remainder 3