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If it is divided by a fraction or a decimal. Like 1/5 or .986

Q: Is it possible for the remainder in a division problem to be greater than the divisor?

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If the remainder were greater than the divisor, you'd be able to take another divisor out of it.

The remainder can be greater than the divisor when the dividend is significantly larger than the divisor. In division, the remainder is the amount that is left over after dividing the dividend by the divisor. If the dividend is much larger than the divisor, it is likely that the remainder will also be larger than the divisor.

The remainder of two positive integers can be calculated by first dividing one number (the dividend) by the other (the divisor) using integer division (ignoring any fractional component). Multiply this quotient by the divisor, then subtract the product from the dividend. The result is the remainder. Alternatively, while the dividend remains greater than the divisor, subtract the divisor from the dividend and repeat until the dividend is smaller than the divisor. The dividend is then the remainder.

No.

Because if the remainder is greater, then you could "fit" another divisor value into it. if they are equal, then you can divide it easily. Thus, the remainder is always lower than the divisor.

Related questions

If the remainder were greater than the divisor, you'd be able to take another divisor out of it.

The remainder can be greater than the divisor when the dividend is significantly larger than the divisor. In division, the remainder is the amount that is left over after dividing the dividend by the divisor. If the dividend is much larger than the divisor, it is likely that the remainder will also be larger than the divisor.

If the remainder is greater than the divisor then you can divide it once more and get one more whole number and then have less remainders.

The problem would not end

Yes, provided the divisor is greater than 5.

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.

It must be less else you have not divided properly; you could divide again 1 or more times!If the remainder is equal to the divisor (or equal to a multiple of the divisor) then you could divide again exactly without remainder. If the remainder is greater but not a multiple of the divisor you could divide again resulting in another remainder.E.g. Consider 9/2. This is 4 remainder 1. Let's say our answer was 3 remainder 3; as our remainder "3" is greater than the divisor "2" we can divide again so we have not carried out our original division correctly!

The remainder is less than the divisor because if the remainder was greater than the divisor, you have the wrong quotient. In other words, you should increase your quotient until your remainder is less than your divisor!

The remainder of two positive integers can be calculated by first dividing one number (the dividend) by the other (the divisor) using integer division (ignoring any fractional component). Multiply this quotient by the divisor, then subtract the product from the dividend. The result is the remainder. Alternatively, while the dividend remains greater than the divisor, subtract the divisor from the dividend and repeat until the dividend is smaller than the divisor. The dividend is then the remainder.

No.

Because if the remainder is greater, then you could "fit" another divisor value into it. if they are equal, then you can divide it easily. Thus, the remainder is always lower than the divisor.

Because if the remainder is greater, then you could "fit" another divisor value into it. if they are equal, then you can divide it easily. Thus, the remainder is always lower than the divisor.