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Compilation of tips and tricks to deal with remainders.

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03 Apr 2010, 13:08
Hi
Please can u explain me more about 4, 5 and 6. I don't understand when u say cycle of powers and how to apply the formulas in 5 and 6.

Thanks a lot.

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13 Apr 2010, 23:01
great post... very informative !
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20 Apr 2010, 16:26
excellent post buddy... +1 to you!
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18 May 2010, 07:27
sriharimurthy, you have made a great post thanks,
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24 May 2010, 03:23

Last edited by takshzilabeta on 24 May 2010, 03:26, edited 1 time in total.
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24 May 2010, 03:24
For Remainders part 2

working with composite index problems
http://takshzilabeta.com/blogs/2010/03/08/1175/

Last edited by takshzilabeta on 19 Jun 2010, 06:29, edited 1 time in total.
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25 May 2010, 04:40
can you plz explain how to find remainder of the fractions that result in a recurring decimal? eg. 7/3 = 2.33333 Therefore,is the remainder 3* 0.3333333 = 0.6666666??

will we then consider the answer to be a multiple of 0.67?
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16 Jul 2010, 21:12
I couldn't follow 5 and 6 but 7 is key.

Thanks
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04 Aug 2010, 19:05
Remainders was giving me a lot of trouble, this was very helpful.

5 and 6 isn't intuitive, but essentially you're breaking it down into small pieces (stripping away the multiples of N) to get the remainder. If you work through real numbers both this way and the long way, it might be more helpful, especially if you break down the x and y into factors.
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30 Aug 2010, 13:27
great post. Thanks
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23 Sep 2010, 06:19
Hello,

Can some one please tell me how to solve the below?

what is the remainder of 11^97/ 7

How do we tackle these type of problems?
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23 Sep 2010, 06:26
Fokir wrote:
Hello,

Can some one please tell me how to solve the below?

what is the remainder of 11^97/ 7

How do we tackle these type of problems?

Let a mod b be the remainder when a is divided by b

Note that xy mod b = ((x mod b)*(y mod b)) mod b

11 mod 7 = 4
11^2 mod 7 = 4*4 mod 7 =2
11^3 mod 7 = 2*4 mod 7 =1
11^4 mod 7 = 1*4 mod 7 = 4
... and then the cycle will repeat

So for 11^97 the remainder will be 4
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18 Oct 2010, 03:42
sriharimurthy can you plz explain the same for this using the remainder logic

Q) find the last 2 digits of 842*9512*408*613

It is pretty easy using the logic stated in

LAST DIGIT OF A PRODUCT

Last digits of a product of integers are last digits of the product of last digits of these integers.

For instance last 2 digits of 845*9512*408*613 would be the last 2 digits of 45*12*8*13=540*104=40*4=160=60

Example: The last digit of 85945*89*58307=5*9*7=45*7=35=5?

Can you please explain it using the remainder logic??
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23 Mar 2011, 07:09
Hi Guys,

I've a question regarding the following point (point number 3) mentioned in the topic " Compilation of tips and tricks to deal with remainders":

If a number leaves a remainder ‘r’ (the number is the divisr) all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor. If the value of ‘r’ is greater than the value of the factor, then we have to take the remainder of ‘r’ divided by the factor to get the remainder.

For example: 1044/33...........remainder is 21

As per my understanding, the factors of 33 i.e 3 and 11 should have the same remainder...

Since the remainder is greater than the factors, the remainder wud be 21/3 (reaminder is 0) and 21/11 (remainder is 10)....the two remainders (o and 10) r diffferent..Can someone pls explain this..is my understanding rite?

Regards,
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24 Mar 2011, 09:37
anuu wrote:
Hi Guys,

I've a question regarding the following point (point number 3) mentioned in the topic " Compilation of tips and tricks to deal with remainders":

If a number leaves a remainder ‘r’ (the number is the divisr) all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor. If the value of ‘r’ is greater than the value of the factor, then we have to take the remainder of ‘r’ divided by the factor to get the remainder.

For example: 1044/33...........remainder is 21

As per my understanding, the factors of 33 i.e 3 and 11 should have the same remainder...

Since the remainder is greater than the factors, the remainder wud be 21/3 (reaminder is 0) and 21/11 (remainder is 10)....the two remainders (o and 10) r diffferent..Can someone pls explain this..is my understanding rite?

Regards,
Anu

33 and factors of 33 will have the same remainder provided the remainder is not greater than any of the factors.

If any of the factors of the divisor (33) are smaller than the remainder of 21, those factors will have different remainders.

Your understanding of how to calculate the remainders for smaller factors is correct.
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24 Mar 2011, 09:40
Also, just wanted to add that these tricks are super helpful. I'd never have thought of these otherwise.
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28 Mar 2011, 08:11
Thanks!! for the quick reply. you got my point cleared.

Yeah, these tricks r very handy. Thks!! to Sriharimurthy for sharing it.
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24 Apr 2011, 11:08
sriharimurthy wrote:
h2polo wrote:
Here is another important property about reminders that everyone should understand:

If you take the decimal portion of the resulting number when you divide by "n", and multiply it to "n", you will get the remainder.

For example, 8/5 = 1.6

.6 * 5 = 3

Therefore, the remainder is 3.

This is important to understand for problems like the one below:

If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45

Good one h2polo.. Added it to the list.

+1 to you!

+1
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06 May 2011, 00:29
hi thanks a lot for these tips on remainders. Kudos to u.
I like the way u easily solved the 2 Remainder questions in some other posts. Could you also provide the links where you have solved questions involving the cycle of powers so as to get a better idea of that too.

thankssss
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08 May 2011, 03:34
Re: Compilation of tips and tricks to deal with remainders.   [#permalink] 08 May 2011, 03:34

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