For example 5 divided by 3 can be expressed as:
- A Fraction: 5/3
- A decimal: 1.67
- A quotient and an integer remainder: 5 divided by 3 gives a quotient of 1 with a remainder of 2.
The quotient is the whole number result of division i.e. the quotient of 5/3 is 1 and the quotient of 11/2 is 5.
A remainder is what’s left after the arithmetic action of division.
In the above example, the multiple of 3 that is nearest to, but still smaller than five is 3. The distance between 3 and 5 is the remainder of 2.
Note that when 8 is divided by 3, the remainder – the distance between 8 and the nearest multiple of 3 (which is 6) – is still 2.
Remainder is defined as the distance (in units) from the dividend to the nearest multiple of the divisor that is smaller than the dividend.
Let’s try a simple example:
Which of the following is the remainder when dividing 13 by 5?
The nearest multiple of 5 that is smaller than 13 is 5 times 2 = 10. The distance between 10 and 13 is 3.
Note that since the remainder is measured from the nearest multiple of the divisor, the greatest possible remainder is one less than the divisor. The remainder can never be equal to, or greater than the divisor. In the above example, a remainder of “8″ when dividing by 5 is illogical, because 8 is greater than 5, and includes another multiple of 5 closer to the dividend.
Likewise, dividing 10 by 5 does not give a remainder of 5 because 10 itself is the nearest multiple of 5 that is closest to itself, giving a remainder of zero (i.e. no remainder).
Identifying remainder problems in the GMAT is easy – the question uses the word remainder (Duh!). GMAT problems involving remainders can usually be solved easily by Plugging in numbers that fit the problem.
Let’s try another example:
When x is divided by 5, the remainder is 3. When y is divided by 5, the remainder is 4. What is the remainder when x + y is divided by 5?
Plug in numbers for x and y that fit the problem: x = 8 and y = 9. These numbers give a remainder of 3 and 4, respectively when divided by 5.
x+y = 8+9 = 17. When 17 is divided by 5, the closest multiple of 5 is 15, and the remainder is 2.
You can plug in a different set if you’re not sure. Try x = 13 and y = 14 – the remainder of x+y will still be 2 when divided by 5.
Note that you could have easily just plugged in the remainders themselves: x = 3 and y = 4 would’ve also worked. The nearest multiple of 5 that is smaller than 3 is zero; the distance between zero and 3, is 3, so the remainder when dividing 3 by 5 is indeed 3. The same goes for 4: the remainder when dividing 4 by 5 is 4 (the difference of 4 from zero). Thus, x = 3 and y = 4 are also valid plug ins. How’s that for an easy calculation?
Where plugging in is difficult to use because the question requires large numbers, use the following equation:
For any integer i divided by another integer d
i = quotient·d + remainder
Thus, when i is divided by 5 the remainder is 3, i can be expressed as the equation:
i = 5x + 3
Here x is the quotient.
This form also supports plugging in: plug in values for x, and you will find the corresponding values of i:
If x = 0 –> i = 5⋅0 + 3 = 3
If x = 1 –> i = 5⋅1 + 3 = 8
If x = 2 –> i = 5⋅2 + 3 =13
and so on.
Remainder is the distance (in units) from the dividend to the nearest multiple of the divisor that is smaller than the dividend.
Now try solving this question:
If i, a and b are integers, is 4(3b + 2) = 5a?
(1) If i is divided by 5 the quotient is a and the remainder is 3
(2) If i is divided by 12 the quotient is b and the remainder is 11