Only itself and one because 113 is a Prime number
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A factor can divide into its multiple evenly.
2 and 60 will divide evenly into 120.
If there is a remainder of 8 when they divide into 113, then they must be factors of 113 - 8 = 105, but not factors of 113, and greater than 8: The factors of 105 are: 1, 3, 5, 7, 15, 21, 35, 105 The factors of 113 are: 1, 113 Thus the 3 numbers which divide 113 with a remainder of 8 are 15, 21, 35, 105 (oops, there are 4 numbers which satisfy the criteria - I never could count...)
multiples of 193
the answer is 14 two times