Q: Why is 2 the largest possible remainder when you divide a whole number by 3?

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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.

The largest number that will divide into 28 42 and 70 with no remainder is 14.

If you divide by 8, the remainder can be any number from 0 to 7.

14. The largest possible number for a remainder is 1 less than the divisor.

The largest possible remainder when dividing by any number N is N-1.

The largest possible number for a remainder is 1 less than the number of the divisor, so it is 5.

No.

The largest remainder would be 8, because if it were 9 you could divide the number once more. The largest remainder you can have is always one less than what you're dividing by. So if you're dividing by 10, your largest remainder is 9. If you're dividing by 100, it's 99. And so on.

The remainder can be any number between 1 and 8 .

24

6

If you mean as in a whole number then it is 7

The largest remainder will be one less than the divisor. 7 - 1 = 6.

24. The largest remainder is always one less than the number by which you are dividing (the divisor).

81 is.

The GCF is 18.

11

As a whole number it is 14

Eleven.

40

It is: 40

It is 22.

If you take the number 969 and subtract the 9 remainder you get 960. Because the number is even you can divide it by 2. 969/2 = 480R9. :)

The largest [integer] remainder is 10. If the remainder was any more you would get one (or more) lots of 11.

The largest number that will divide evenly into 64 with no remainder is 64.

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