mathematics
2
9. The divisor must be greater than the remainder. A 1 digit divisor that is greater than 8 can only be 9.
That's not possible. The largest single-digit number by which you might divide is 9. And, by definition, the remainder is always LESS than the number by which you divide. Thus, the largest remainder you can get, when you divide by 9, is 8.
23 divide 3974
81 is.
Regardless of the dividend (the number being divided), no divisor can produce a remainder equal to, or greater than, itself..... dividing by 4 cannot result in a remainder of 5, for example, Therefore the only single-digit number which can return a remainder of 8 is 9. 35 ÷ 9 = 3 and remainder 8
Divide the two-digit number by the one-digit number. If the remainder is zero then the 2-digit number is a multiple and if not, it is not.
im sorry but i can t figure it out
You can't have a remainder of 6 when you divide by 2! JHC!
0.28235294117647
20 if you divide by 17. 19 if you divide by 16. 18 if you divide by 15, 17 if you divide by 14. And so on. In fact any number from 10 to 99. That is, every two digit number.
5, or .8 with a three repeating.