Q: If a division problem has divisor of 9 what is the greatest possible remainder?

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The greatest integer remainder for a division sum with a divisor of 63 would be 62 - for a number one fewer than an integer multiple of 63 - for example, 125/63 = 1 remainder 62.

62

16

7

75

24. It is always one less than the divisor.

12-1, that is, 11.

456y67783

The greatest possible remainder for 76 is to be divided by 2 leaving the result of 38. Thank you!

What is the largest remainder possible if the divisor is 10

the parts of division problem are : dividend , divisor , quotient and remainder . where : dividend = quotient * divisor + remainder

The greatest possible remainder is divisor is just less than 28. If the dividend is an integer, then it is 27. But if the dividend is any real number, the greatest possible remainder is 27.999.... (recurring).Strictly speaking, though, the decimal can only be "nearly recurring". This is because 27.999 recurring = 28 and the remainder cannot be 28.The greatest possible remainder is divisor is just less than 28. If the dividend is an integer, then it is 27. But if the dividend is any real number, the greatest possible remainder is 27.999.... (recurring).Strictly speaking, though, the decimal can only be "nearly recurring". This is because 27.999 recurring = 28 and the remainder cannot be 28.The greatest possible remainder is divisor is just less than 28. If the dividend is an integer, then it is 27. But if the dividend is any real number, the greatest possible remainder is 27.999.... (recurring).Strictly speaking, though, the decimal can only be "nearly recurring". This is because 27.999 recurring = 28 and the remainder cannot be 28.The greatest possible remainder is divisor is just less than 28. If the dividend is an integer, then it is 27. But if the dividend is any real number, the greatest possible remainder is 27.999.... (recurring).Strictly speaking, though, the decimal can only be "nearly recurring". This is because 27.999 recurring = 28 and the remainder cannot be 28.

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

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

One less than the divisor, so 6.

7

44

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

The answer depends on what the divisor is.

The greatest integer remainder is 7 but otherwise, 7.999... .

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

The remainder CAN'T be greater than the divisor, not if you do the division correctly.

37

You do not invert it. However, you can convert the remainder to a decimal by carrying out a long division of the remainder divided by the original divisor. For example, 13/3 = 4r1 Then, long division of the remainder (=1) by the divisor (=3) gives 0.33.... which is the converted remainder. The full quotient, in decimal form is 4.33...