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Q: What is the largest possible remainder for a math problem with 8 as the divisor?

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456y67783

Apparently, you're only using whole numbers in your division. In that case, the largest possible remainder is two (2).

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.

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...

The remainder can be: 0,1,2,3.

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

the divisor can not have that number going into the dividend anymore.

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