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quotent X divisor + remainder = dividend

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Q: Divisor and remainder is given find the dividend?
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How do we find remainder of two numbers by using addition subtraction multiplication and division?

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.


How do you find the answer to a division problem?

Divide the divisor into the dividend which will result as a quotient and sometimes having a remainder


What is 71 over 8 as a mixed number?

To find the mixed number you need to first divide to find the quotient and remainder. So 71 over 8 has a quotient of 8 and remainder 7. So the general way of writing a mixed number is dividend over divisor = quotient (remainder over divisor) dividend/divisor = quotient remainder/divisor) So 71 over 8 = 8 7/8.


How do you find the dividend with a divisor 7?

If there is no remainder, you can use the relation:dividend = divisor x quotient If you ONLY know the divisor, you don't have enough information; though you can make up any number for the quotient, and multiply them together to get the dividend.


How do you find the divisor if you know the quotient is 41 and the dividend is 1681?

0.0244


How do you use the reciprocal of the divisor to find the answer to a division problem?

Multiply the reciprocal of the divisor by the dividend.


How do you find a missing dividend in a division problem?

The three parts to a division problem are: Dividend, Divisor, and Qoutient. To calculate the value of any of the terms, two of the terms need to be known values. To calculate the dividend, multiply the quotient by the divisor.


What is the greatest common factor of 25 and 63?

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.)


How can you use multiplication to check a division problem?

we can multiply the divisor & the quotient to find the dividend


Find the quiotent when the divisor is 25 and the dividend is 150?

0.1667


What is a Java program that finds the greatest common factor of two integers?

The easiest way to find the greatest common denominator of two integers with a computer program is to use the Euclidean algorithm. Of the most popular methods of finding the GCD of two numbers, the Euclidean algorithm does it with the least amount of work and requires the least amount of code.In order to understand the Euclidean algorithm, you'll need to know a few division terms:The dividend is the number to be divided.The divisor is the number being divided by.The quotient is the number of times the divisor divides into the dividend.The remainder is the amount "left over" when the divisor cannot go into the dividend an integral number of times.18A divided by 12B gives a quotient of 1C and a remainder of 6D. A is the dividend, B is the divisor, C is the quotient, and D is the remainder.The Euclidean algorithm works like this:Check if either of the two integers is 0. If so, there is no solution (Ø), as a number cannot share a GCD with zero. Besides, division by zero is a big no-no.Check if either of the two integers is 1. If so, 1 is the GCD.Divide the larger of the two integers by the smaller.Divide the divisor of the previous division operation by the remainder of the previous operation.Repeat step four until the remainder equals zero. When the remainder equals zero, the divisor of the last operation is the GCD.If you still don't get it, try looking at the Euclidean algorithm in action:Find the GCD of 84 and 18.Check to see if either 84 or 18 is equal to 0. Nope. Continue on...Check to see if either 84 or 18 is equal to 1. Nope. Continue on...Since 84 is larger than 18, divide 84 by 18. Quotient is 4, remainder is 12.Take the divisor of the last operation (18) and divide it by the remainder of the last operation (12). Quotient is 1, remainder is 6.Take the divisor of the last operation (12) and divide it by the remainder of the last operation (6). Quotient is 2, remainder is 0.When the remainder is 0, the divisor of the last operation is the GCD. So the GCD in this case is 6.You should now have a good grasp of how the Euclidean algorithm works. Now we need to turn it into code. We'll need three variables, all of them integers:int divisor, dividend, remainder;The purpose of the variables is self-explanatory. Next, we need to make a few decisions. We need to decide if the dividend or the divisor is 0. If that test is passed, then we need to decide if the dividend or the divisor is 1. If that test is passed, then we need make sure that dividend is larger than divisor.if(dividend 1) {printf("The GCD is 1.\n");}// Make sure the dividend is greater than the divisor.if(divisor > dividend) {remainder = dividend;dividend = divisor;divisor = remainder;}// Calculate the GCD.while(remainder != 0) {remainder = dividend % divisor;dividend = divisor;divisor = remainder;}// Display the answer to the user.printf("The GCD is %i.\n", dividend);}And the GCD lived happily ever after. The end.


Find a number which when subtracted from 3214682 will make it exactly divisible by 487?

3214682/487 gives 6600 as quotient 482 remainder. Dividend-remainder=divisor*quotient 3214682-482 gives 3214200 which is divisible by 487. 482 can be subracted there are more possibility