Ambiguity in remainder based problem

[xerocoool]

Q) a,b,c are multiples of 15 and a<b<c
Quantity A: Remainder when b is divided by c
Quantity B: Remainder when (b+c) is divided by a

Case 1 : 15 < 30 < 45
Quantity A : 45 R
Quantity B : 0 R

Case 2 : 15 < 60 < 150
Quantity A : 60 R
Quantity B : 0 R

Clearly A is greater (which is the answer given in the book)

But what if we consider,
a<b<c
a = - 45
b = - 30
c = - 15
Quantity A : 0 R
Quantity B : 0 R
hence (C)

and the final answer (D). Is it the right way of going about ?

[asishkm]

Hi, I think you are spot on in your analysis.

Since it is not mentioned that 0<a<b<c, it can be assumed that they are negative as well. Hence the correct answer would be (D)



However, it is my view that ETS won't ask anything regarding division or remainders with respect to negative integers, especially since there is no consensus on what is taken to be the remainder when the divisor is negative. Most of such ambiguous questions will be explicitly stated as to whether negatives are allowed or not. For example, even in the case of divisors and multiples, questions will state if it means positive divisors/multiples or non-zero or non-negative etc. so that there is no ambiguity.

For example if you divide (-11)/(-5), then some sources state the remainder must be the one that has the least absolute value which in this case would be -1 , -11 = (2)*(-5) + (-1)
Some other sources state that the remainder should be 4 because the remainder is defined to be 0 <= r < |d| in which case -11 = (3)*(-5) + 4

[xerocoool]

Hey,

Yes I believe so, since there are no constraints we can consider negative multiples. (I haven't come across many sums where negative multiples have been tested, so maybe)

Regarding remainders,

I am not really sure about the absolute value thing. I just make sure the remainder stays within the range 0<= R < N (Remainder should be strictly greater than or equal to 0)

For ex.
-16/3
Quotient will be -6
Remainder : 2

2) Also, Can we not consider -11/-3 as just 11/3 ? (I am not really sure on this) and solve as 2 positive integers ?

[asishkm]

If the divisor is positive, then there is no issue regarding the remainder which will definitely lie between 0 and the divisor.

However, issues seem to arise when the divisor is negative. Do you want the remainder to lie between 0 and the magnitude (absolute value) of the divisor in which case it will be positive? or do you intend it to lie between 0 and the divisor (in which case it will be negative)? or do you want it to be the one which is closest in distance (or separation) from 0?

2) For (-11)/(-3), if you treat it as 11/3, you'll get a quotient of 3 and a remainder of 2

i.e. 11 = (3)(3) + 2
which implies that -11 = (3)(-3) + (-2) (multiplying both sides by -1) i.e. remainder is -2

If one sticks to the "fact" that the remainder must be non-negative, then obviously the method of considering -11/(-3) to be equivalent to 11/3 becomes flawed. In this case -11 = (4)(-3) + 1 which leads to a remainder of 1.

The difference between the two ways of calculating the remainder is more apparent if you take (-14)/(-3), then you can express it in the following two ways:
-13 = (4)(-3) + (-1): remainder = -1 (this is also the remainder obtained when treating it as 13/3 and then multiplying both sides by -1)

-13 = (5)(-3) + 2: remainder = 2

- If you consider the definition of the remainder to be the one with the least magnitude (distance from 0), then it should be -1
- If you consider the definition of the remainder to lie between [0, |b|) where b is the divisor, then it should be 2


Further, the ETS official guide only states that:

In general, when a positive integer a is divided by a positive integer b, you first find the greatest multiple of b that is less than or equal to a. That multiple of b can be expressed as the product qb, where q is the quotient. Then the remainder is equal to a minus that multiple of b, or r = a - qb, where r is the remainder. The remainder is always greater than or equal to 0 and less than b.

Even the official guide refrains from including any negative integers while dividing.
So, I would not worry too much about division or remainders of negative integers, since there is no "single correct" definition in such case.

[xerocoool]

Hey,

Thanks for the detailed insights.

1) Understood the difference between finding remainders for (11/3) and (-11/-3). Will make a note of it.

2) To be honest, I had no idea about the second definition of remainders "the least magnitude from 0" :: better to stick to the ETS one "0 <= R < |b|"


-Xero

[tommywallach]

Just to make sure this is intensely clear: There has never been a single question that considers negative multiples/divisors of a number, and CERTAINLY no question that then considers the consequences of this for remainders. All of this conversation is interesting, but it has nothing to do with anything that will show up on the GRE.

: )

-t

[xerocoool]

haha :)

Thanks for the update Tommy !

Xero

[asishkm]

Thanks Tommy for the update!

Cheers.

[tommywallach]

Ha! Glad to be of help (...which I gave absolutely none of!).

-t