If n is a negative integer and q is a positive integer, then there is

Given

• n → negative
• q → positive
• r → positive

Inference

• m → negative

As r is positive, to obtain the value of m, we will have to take the greatest integer lower than or equal to the value of (\(\frac{-100}{30}\)), therefore \(m = -4\)

\(-100 = 30 * (-4) + r\)

\(r = 120 - 100 = 20\)

Option E

 

­I do not get it. If I divide -100 by 30, it divides 3 times and gives me the reminder -10, doesn't it? 

­You cannot have negative remainders. Remainders are always non-negative.

To convert negative remainder to positive add that to the divisor.
Hereit is 30+(-10) or 20

 

dividend = the integer being divided
divisor = the integer being divided into the dividend
quotient = the result after division
remainder = the portion of the dividend that REMAINS after division

Putting it all together:
dividend = (divisor)(quotient) + remainder

When the division of two integers yields a REMAINDER, the reason is that the product in blue is less than than the dividend, so part of the dividend REMAINS.

In the case at hand:
-100 = dividend
30 = divisor
Thus:
-100 = (30)(quotient) + R

Here, a quotient of -3 will make the product in blue GREATER than the dividend in red, so it would be illogical to claim that part of the dividend remains.
For this reason, the quotient here will generally be thought to be not -3 but -4, so that the product in blue is LESS than the dividend of -100:
-100 = (30)(-4) + R
R = 20
 ­

n = mq + r

n = -ve, q = +ve, where 0 ≤ r ≤ q.

i.e. m MUST be -ve

What is remainder when -100 is divided by 30?



When -100 is divided by 30 then Quotient is -3 and remainder is -10

but -10 remainder is same as remaindr 30-10 = +20

Answer: Option E

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­

Are there any similar questions to practice?

­Check PS Remainders Questions.

Is this really a remainder question? You can really just fill in the missing values.

N=mq+r
N has to be -100
M is only one negative integer
q has to be 30
r has to be a integer that is positive -> that eliminates A) and B) already

New equation is: -100=(M)(30)+R
Literally just plug C,D,E into R and see if you can make equation work.

Only E) 20 works because M will equal -4

-100=(-4)(30) + 20

 

­
The toughest part of this question is keeping track of the signs of the variables. I would first make sense of this:

there is only one negative integer m such that n = mq + r, where 0 ≤ r ≤ q (n is negative, q positive, m negative and r is positive)

Plug in the simplest values. Say -3 = m*2 + r
m must be negative. If m = -1, r will also need to be negative. Say m = -2 which will give r = 1. Ok.
If m = -3, r = 3 which is greater than q hence not acceptable. So essentially, m is the negative number which gives us a value just smaller than n and r is positive to bring the balance on the right hand side. 

Similarly then, - 100 = m*30 + r
Till m = -3 will not work since it will give a value greater than - 100.
m must be -4 so that - 100 = - 120 + r
r = 20

Answer (E)