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62. One less than the divisor.

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Q: What is the largest remainder possible if the divisor is 63?

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62

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.

To determine the remainder, you would take 63 and see how many times your divisor fits into it. That will give you a number, which when multiplied by the divisor will be less than 63, and smaller than the divisor. Subtract the result of your divisors times your quotient from 63, and that number is the remainder.

Then divide the remainder again by the divisor until you get a remainder smaller than your divisor or an remainder equal to zero. The remainder in a division question should never be larger than the "divisor", but the remainder often is larger than the "answer" (quotient). For example, if 435 is divided by 63, the quotient is 22 and the remainder is 57.

Why not use the Euclidean Algorithm and find out? Divide 63 by 25, and you get a remainder of 13. (The quotient is not important.) Now the divisor of the last division problem becomes the dividend, and the remainder becomes the divisor - that is, we divide 25 by 13 this time. We get a remainder of 12. Divide 13 by 12, and you get a remainder of 1. Divide 12 by 1, you get no remainder. Therefore, this last divisor, 1, is the greatest common factor (or divisor) of the original two numbers. (As a side note, because the gcf is 1, that means those two numbers are what's called relatively prime.)

The Greatest Common Divisor of 35, 63 is 7.

7

The Greatest Common Divisor (GCD) for 18 63 is 9.

The greatest common divisor of 63 and 81 is nine (9).

0.0635

1, 3, 7, 9, 21, 63.

76.1111

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