2 or 8.
No.
180
Any multiple of 60.
Out of that list, 2, 3 and 6.
The number would be (7x23456) + 1 = 164193
23457 divided by 23456 leaves a remainder of 1, and it is divisible by 7. So that is one of infinitely many possible answers.
23456-56789 = -33333
23456 x 10 = 234560..If you are looking for the smallest possible number divisible by 23456, however, that's another ball game. Then you'd have to break both numbers into prime factors:10 = 2 x 523456 = 2 x 2 x 2 x 2 x 2 x 733 (yes, 733 is a prime number)Then you need to make sure that every prime factor from each of these numbers are "represented" in the number that'll be divisible by both 10 and 23456. However prime factors from different numbers can overlap (here the 2 from 10 and one of the 2's from 23456 are not written as two separate factors, but as one).Thus:2 x 5 x 2 x 2 x 2 x 2 x 733 = 10 x 11728 = 117280.Note that as we removed one 2 as a factor, the number is now half of my first and most simple suggestion. This number, however, should be the absolute smallest number divisible by both 10 and 23456.
To find the least number divisible by 23456 with a remainder of 1, you need to find the least common multiple (LCM) of 23456 and the number 1 more than it. The LCM of two numbers is the smallest number that is a multiple of both numbers. In this case, you would calculate the LCM of 23456 and 23457. The LCM can be found by multiplying the two numbers and then dividing by their greatest common divisor.
23456 + 758 = 24214
23456 + 654321 = 677,777