For each of the following, what are the possible remainders

[agkp1984]

For each of the following, what are the possible remainders upon division by 4?
(D) The product of two consecutive odd integers.

The answer is 3.

My doubt is if we consider 2n-1 and 2n+1 as two consecutive odd integers then the remainder is 1.
(2n-1)(2n+1) = 4n^2 - 1.
However if we try actual numbers the answer comes out to be 3. Can someone please explain where I am going wrong? The answer explanation given in the Guide does not cover the scenario I have provided.

Source: The Number Properties Guide, 4th Edition, chapter 11, Pg-153. Q-5D

Thanks and kind regards,
Arun

[mithunsam]

How did you arrive at "2n-1 and 2n+1 as two consecutive odd integers then the remainder is 1"?

(2n-1)(2n+1) = 4n^2 - 1

4n^2 - 1 = 4I - 1, where I is any +ve integer (n^2 can only be +ve).

Now consider 4I (this is a multiple of 4)

But, 4I = 4(I-1) + 4

Subtract 1 from both sides
4I - 1 = 4(I-1) + 4 -1
That is, 4I - 1 = 4(I-1) + 3

In otherwords, 4n^2 - 1 = 4(I-1) + 3, where (I-1) is either 0 or a +ve integer

So, the reminder will always be 3. Does that make sense?

[agkp1984]

Mithunsam,

Thanks for the quick response. However I am still not clear with your approach.

Let us consider the example of guide
(2n+1) and (2n+3)
= (2n+1)(2n+3)
= 4n^2 + 8n + 3 Hence the remainder 3. However if I follow your approach then,
=(4n^2 + 8n + 4) - 4 + 3
=4(n^2 + 2n +1) - 1
=4(n+1)^2 - 1 Hence the remainder would come to 1.

I am not clear why you performed the addition and subtraction of 4. So if you could elaborate on why you performed this additional step as it is not required while solving similar examples as indicated above, then it would be clear to me.

Regarding how 2n-1 and 2n+1 are consecutive odd integers. If I consider 2n as any even integer then 1 number less than 2n and 1 number more than 2n are its consecutive integers and they are both odd. Hence 2n-1 and 2n+1 are consecutive odd integers.

Thanks and kind regards,
Arun

[messi10]

Hi,

Whenever a number is divided by a positive integer "n" then the range of possible remainders is from 0 to n - 1.

I am assuming that this is what mithunsam was trying to demonstrate. Let's forget the algebraic approach for a minute here. Let's say I wanted to divide 7 by 4. Its easy to see that the remainder is 3 in this case. We do not say that it goes 4 x 2 = 8 and the remainder is -1.

When you start considering negative dividends and divisors, this gets slightly complicated. Let's say take an example of 2n+1 and 2n-1. Let's say n = 0:

(2n+1) x (2n-1) = -1 x 1 = -1

Now we have -1 divided by 4. Intuitively we may say that 4 times 0 gives us 0 and the remainder is -1. However, this means that we are breaking the law of simple division. We must divide by a number that is either equal to or smaller than the dividend. So mathematically 0 is greater than -1. So when dividing -1 by 4, we say that 4 x -1 =-4 leaving the remainder as 3. This can be written as:

-1 = (-1 x 4) + 3

Hope this helps

Regards

Sunil

[mithunsam]

Dividend = Quotient*Divisor + Reminder
You need to convert your number to this format. You are keeping it as Quotient*Divisor - Reminder, which is wrong.

Now, let us consider your example 4(n+1)^2 - 1.

To find the reminder when divide by 4, we have to rewrite this number to Q*4 + R format

We can re-write 4(n+1)^2 - 1 as 4I - 1, where I is some +ve integer. ---> Step 1
[I is always +ve as (n+1)^2 is +ve]

But 4I is 4(I-1) + 4. ---> Step 2
[For example
16 = 4(4) = 4(4-1) + 4 = 4(3) +4
16 = 4(I) = 4(I-1) + 4, where I = 4

Similarly
20 = 4(5) = 4(5-1) + 4 = 4(4) +4
20 = 4(I) = 4(I-1) + 4, where I = 5]
Subtract 1
4 I - 1 = 4(I-1) + 4 -1 = 4(I-1) + 3 ---> Step 3

Now re-substitute I = (n+1)^2 into 4I - 1 ---> Step 4
4 (n+1)^2 - 1 = 4(I-1) + 3
(Your example) = Q*4 + R => R = 3 (Q=I-1)

If you still have question, please mention the step numbers above, which you aren't clear. I would try to explain that.

[jnelson0612]

Wow, mithunsam and Sunil both weighing in on a question! This thread is fantastic--great work!