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...)
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To find how many times the number 3 is in 113, you can divide 113 by 3. Doing the calculation, 113 divided by 3 equals approximately 37.67. This means that 3 fits into 113 a total of 37 full times, with a remainder.
52 thats your answer :)
When you divide 24 by 7 the remainder is 3.
If you take any four consecutive numbers and divide them by 3, the remainders are as follows: 9/3 = 3 10/3 = 3 remainder 1 11/3 = 3 remainder 2 12/3 = 4 Therefore, the highest remainder you can have by dividing a whole number is 2.
You show your remainder by circling the equal group of numbers and then the remainder stays the same you don't circle it.