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You are watching: Quotient and remainder of negative numbers

I might have sworn that I had actually discussed negative remainders on mine blog but the other day ns was searching for a short article discussing it and also much together I would try, I could not find one. Ns am a tiny surprised because this ide is quite useful and also I should have actually taken it in information while mentioning divisibility. Though us did have actually a fleeting conversation of that here.

Since us did miss out on it, us will talk about it in detail today yet you should review the connect given over before us proceed.

Consider this: when n is divided by 3, it pipeline a remainder 1.

This means that as soon as we divide n balls in teams of 3 balls each, we space left with 1 ball.

This also way that n is 1 more than a multiple of 3. Or, that also way that n is 2 much less than the following multiple the 3, no it?

Say, n is 16. Once you split 16 balls into teams of 3 balls each, you get 5 groups of 3 balls each and there is one ball leftover. N is 1 more than a many of 3 (the multiple being 15). But we can additionally say the it is 2 less than the following multiple the 3 (which is 18). Hence, the negative remainder in this case is -2 i beg your pardon is indistinguishable to a confident remainder that 1.

Generally speaking, if n is split by m and also it pipeline a remainder r, the negative remainder in this instance is -(m – r).

When n is divided by 7, it leaves a remainder of 4. This is tantamount to a remainder of -3.

n is 3 more than a many of m. The is likewise 2 less than the following multiple of m. This way m is 5.

This ide is really useful to us sometimes, particularly when the divisor and also the remainder are big numbers.

Let’s take it a inquiry to see how.

Question 1: What is the remainder when 1555 * 1557 * 1559 is split by 13?

(A) 0

(B) 2

(C) 4

(D) 9

(E) 11

Solution: due to the fact that it is a GMAT concern (a concern for which us will have no calculator), multiply the 3 numbers and also then dividing by 13 is absolutely the end of question! There needs to be an additional method.

Say n = 1555 * 1557 * 1559

When we divide 1555 by 13, we get a quotient of 119 (irrelevant to our question) and also remainder the 8. So the remainder once we division 1557 by 13 will certainly be 8+2 = 10 (since 1557 is 2 more than 1555) and when we divide 1559 by 13, the remainder will be 10+2 = 12 (since 1559 is 2 more than 1557).

So n = (13*119 + 8)*(13*119 + 10)*(13*119 + 12) (you can pick to overlook the quotient and just create it together ‘a’ due to the fact that it is irregularity to ours discussion)

So we need to uncover the remainder once n is separated by 13.

Note that once we multiply these factors, every terms we obtain will have 13 in them other than the last term which is acquired by multiplying the constants with each other i.e. 8*10*12.

Since all various other terms are multiples that 13, we have the right to say that n is 8*10*12 (= 960) an ext than a many of 13. There space many more groups the 13 balls that we can kind out of 960.

960 split by 13 offers a remainder of 11.

Hence n is actually 11 an ext than a multiple of 13.

We did not use the an adverse remainders principle here. Let’s see how using negative remainders makes our calculations easier here. The remainder the 8, 10 and 12 suggest that the negative remainders are -5, -3 and -1 respectively.

Now n = (13a – 5) * (13a – 3) * (13a – 1)

The critical term in this instance is -5*-3*-1 = -15

This way that n is 15 much less than a multiple of 13 i.e. In reality 2 less than a many of 13 because when you go ago 13 steps, girlfriend get one more multiple of 13. This gives us a an adverse remainder the -2 which way the optimistic remainder in this case will it is in 11.

Here us avoided some large calculations.

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I will leave you now with a question which you should shot to fix using an adverse remainders.

Question 2: What is the remainder as soon as 3^(7^11) is split by 5? (here, 3 is increased to the strength (7^11))

(A) 0(B) 1(C) 2(D) 3(E) 4

Hint: I resolved this inquiry orally in a few secs making use of cyclicity and an adverse remainders. Don’t obtain lost in calculations!

Karishma, a computer Engineer through a keen attention in alternate Mathematical approaches, has mentored college student in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and also regularly participates in content breakthrough projects such as this blog!