1.
The divisor is the second number in a division problem. For instance 6 / 3 = 2. In this example, the divisor is 3.
If you have a divisor of X, then the largest remainder possible is X-1. This is because if you had one more number in the remainder, it would form a complete count, and the remainder would go away.
In the case of 2 as your divisor, think of the number 11. 11 / 2 has a remainder of 1. However, if you had one more in the remainder, you'd have 2, and that would be a complete division. (Also, the number you have to be 12.) And there would be no remainder.
If you divide by 8, the remainder can be any number from 0 to 7.
Apparently, you're only using whole numbers in your division. In that case, the largest possible remainder is two (2).
If the divisor of a division problem is 4, any number between 0 and 3 (inclusive) can be a remainder for that problem.
If it is divided by a fraction or a decimal. Like 1/5 or .986
the divisor can not have that number going into the dividend anymore.
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If you divide by 8, the remainder can be any number from 0 to 7.
Apparently, you're only using whole numbers in your division. In that case, the largest possible remainder is two (2).
The remainder is the number that is left over after the initial value has been divided as much as it can. If any numbers greater than 48 were present as a remainder, then these could be divided further into 48. If 48 is present as the remainder, then this can be divided by 48 to give 1, leaving no remainder. Thus, the largest possible remainder if the divisor is 48 is 47.
8 is the greatest possible whole number remainder, eg seventeen divided by nine...
If the divisor of a division problem is 4, any number between 0 and 3 (inclusive) can be a remainder for that problem.
the parts of division problem are : dividend , divisor , quotient and remainder . where : dividend = quotient * divisor + remainder
If it is divided by a fraction or a decimal. Like 1/5 or .986
No, cause the remainder might be bigger than divisor.
quotient,divisor, and dividend and remainder
The problem would not end
what