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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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08 May 2011, 10:31
russty wrote: 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 Hi, Here's a link regarding remainders and cycles of power. I found it very helpful. I guess this link was mentioned somewhere in this forum only.I had stored it in my favorite list..hope u too find it useful. http://takshzilabeta.com/catquant/numb ... rsparti/Anu



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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09 May 2011, 21:55
[/quote] Hi, Here's a link regarding remainders and cycles of power. I found it very helpful. I guess this link was mentioned somewhere in this forum only.I had stored it in my favorite list..hope u too find it useful. http://takshzilabeta.com/catquant/numb ... rsparti/Anu[/quote] Hey thanks Anu!! I must've missed the link if it was mentioned before.



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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13 May 2011, 11:42
Nice post!... it really cleared my doubts in remainders!.. thanks alot



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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13 May 2011, 13:48
Just to clarify, for tip #5, if the numerator is smaller than the denominator then the remainder is simply the numerator? For example, 20/25 R = 20 and 4/13 R = 4?



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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13 May 2011, 14:33
@seoulite Just to clarify, for tip #5, if the numerator is smaller than the denominator then the remainder is simply the numerator? For example, 20/25 R = 20 and 4/13 R = 4?
yes... coz, when you are dividing 4 by 13, quotient will be '0' and reminder will be 4.



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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16 Jun 2011, 04:17
thanks a ton for these tips and tricks.. I have another question which one of my friend gave me.I do not have the way to solve it ((38^16!)^17777 ) mod 17 = ? options: 1 4 15 16 Any quick way to solve this ? thanks! Regards,



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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05 Sep 2011, 22:53
liked the last part of the post. Thanks



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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07 Sep 2011, 10:30
getmba wrote: Excellent job sriharimurthy! Very helpful post on remainders. Good initiative. +1. Thanks, sriharimurthy!



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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15 Sep 2011, 19:00
Hey remainders are one of the trickiest topics in the GMAT thanks for the heads up



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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19 Oct 2011, 11:36
Its really good. esp points 5,6,7
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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28 Oct 2011, 19:41
Thanks so much! Great addition to my study notes



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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03 Jan 2012, 04:30
great post. Kudos to you!



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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16 Feb 2012, 02:10
many thanks to you, just when I was almost driven crazy by those "Remainder problem" I found this post. It really saves my day



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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21 Feb 2012, 09:24
sriharimurthy wrote: 3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor. Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5.
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. Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 3 (which is a factor of 21) will be remainder of 5/3, which is 2.
ummm.....this cant be true.....say 147/35....remainder is 1.....7 is a factor of 35....so it should also leave remainder 1 ....but 147 /7 leaves a remainder 0..... am I missing smting?



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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14 Apr 2012, 04:18
Amazing!!!)) I'm happy I'm working on gmat  even if i don't pass it, my knowledge will stay w me! I do Love this forum!



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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04 May 2012, 06:18
sriharimurthy wrote: Hi guys, This is in conjunction with another post which has questions dealing with remainders ( collectionofremainderproblemsingmat74776.html). I'm just trying to put together a list of tips and tricks which we can use to solve these kind of problems with greater accuracy and speed. Please feel free to comment and make suggestions. Hopefully we can add onto this list and cover all sorts of strategies that would help us deal with remainders! Cheers. Please read this first : 1) Take your time with these points. Some of them might be a little difficult to follow in the first reading, but don't give up. The concepts are fairly simple. 2) These tips if mastered will be extremely valuable in the GMAT to help solve a variety of questions not limited specifically to remainders. I have been using them for quite a while now and they have not only helped me improve my accuracy but also my speed. 3) If you have any doubts, please do not hesitate to ask (no matter how stupid you might think them to be!). If you do not ask, you will never learn. 4) Lastly, have fun while trying to understand these tips and tricks as that, according to me, is the best possible way to learn.
All the best!xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx NOTE: Where ever you see R of 'x' it just stands for Remainder of x. 1) The possible remainders when a number is divided by ‘n’ can range from 0 to (n1). Eg. If n=10, possible remainders are 0,1,2,3,4,5,6,7,8 and 9.2) If a number is divided by 10, its remainder is the last digit of that number. If it is divided by 100 then the remainder is the last two digits and so on. This is good for questions such as : ' What is the last digit of.....' or ' What are the last two digits of.....' .3) If a number leaves a remainder ‘r’ (the number is the divisor), all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor. Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 7 (which is a factor of 21) will also be 5. 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. Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 3 (which is a factor of 21) will be remainder of 5/3, which is 2. 4) Cycle of powers : This is used to find the remainder of \(n^x\), when divided by 10, as it helps us in figuring out the last digit of \(n^x\).The cycle of powers for numbers from 2 to 10 is given below: 2: 2, 4, 8, 6 → all \(2^{4x}\) will have the same last digit. 3: 3, 9, 7, 1 → all \(3^{4x}\) will have the same last digit. 4: 4, 6 → all \(4^{2x}\) will have the same last digit. 5: 5 → all \(5^x\) will have the same last digit. 6: 6 → all \(6^x\) will have the same last digit. 7: 7, 9, 3, 1 → all \(7^{4x}\) will have the same last digit. 8: 8, 4, 2, 6 → all \(8^{4x}\) will have the same last digit. 9: 9, 1 → all \(9^{2x}\) will have the same last digit. 10: 0 → all \(10^x\) will have the same last digit. 5) Many seemingly difficult remainder problems can be simplified using the following formula : \(R of \frac{x*y}{n} = R of \frac{(R of \frac{x}{n})*(R of \frac{y}{n})}{n}\) Eg. \(R of \frac{20*27}{25} = R of \frac{(R of \frac{20}{25})*(R of \frac{27}{25})}{25} = R of \frac{(20)*(2)}{25} = R of \frac{40}{25} = 15\)
Eg. \(R of \frac{225}{13} = R of \frac{(15)*(15)}{13} = R of {(2)*(2)}{13} = R of \frac{4}{13} = 4\) 6) \(R of \frac{x*y}{n}\) , can sometimes be easier calculated if we take it as \(R of \frac{(R of \frac{(xn)}{n})*(R of \frac{(yn)}{n})}{n}\) Especially when x and y are both just slightly less than n. This can be easier understood with an example: Eg. \(R of \frac{(19)*(21)}{25} = R of \frac{(6)*(4)}{25} = 24\)
NOTE: Incase the answer comes negative, (if x is less than n but y is greater than n) then we have to simply add the remainder to n. Eg. \(R of \frac{(23)*(27)}{25} = R of \frac{(2)*(2)}{25} = 4.\) Now, since it is negative, we have to add it to 25.\(R = 25 + (4) = 21\) [Note: Go here to practice two good problems where you can use some of these concepts explained : numbers86325.html] 7) 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. [Special thanks to h2polo for this one] Note: Converse is also true. If you take the remainder of a number when divided by 'n', and divide it by 'n', it will give us the remainder in decimal format. Eg. \(\frac{8}{5} = 1.6\)
In this case, \(0.6 * 5 = 3\)
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) 45OA : hello thanks so MUCH for sharing please can anyone explain me the folloxwing formula or give example or actual gmat question where we can apply that the formula is number 5 and 6 From your post here 5) Many seemingly difficult remainder problems can be simplified using the following formula : \(R of \frac{x*y}{n} = R of \frac{(R of \frac{x}{n})*(R of \frac{y}{n})}{n}\) 6) \(R of \frac{x*y}{n}\) , can sometimes be easier calculated if we take it as \(R of \frac{(R of \frac{(xn)}{n})*(R of \frac{(yn)}{n})}{n}\) THANKS BEST REGARDS



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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28 May 2012, 10:12
Thanks for the compilation of such a nice shortcuts Really good post!!
Kudos +1 !!



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Re: Practice these tricks! [#permalink]
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28 May 2012, 10:33
sriharimurthy wrote: R2I4D wrote: Practice sri's wonderful tips and tricks: 4GMAT Home >> GMAT Test Prep Questions >> Number Systems...
(Can't post hyperlinks yet, sorry!)
I was wondering if someone could show me how to do this problem (found at the said site) quicker than my method (detailed below):
8. What is the remainder when the product of 1044, 1047, 1050, and 1053 is divided by 33?
I used sri's trick and found that 1056 is a multiple of 33. This resulted in remainders of (12)(9)(6)(3), respectively. Multiplied together, you get 1944. The remainder of 1944, when divided by 33, is 30, the correct answer.
Is there a quicker way than multiplying (12)(9)(6)(3) out?
Thanks, in advance! Hi, Im glad you found these tips helpful. There is in fact a quicker way to solve it. R of \(\frac{(12)*(3)*(9)*(6)}{33}\) = R of \(\frac{(36)*(54)}{33}\) = R of \(\frac{(3)*(21)}{33}\) = R of \(\frac{63}{33}\) = \(30\) As you can see, you don't really need to do any complex multiplications. Just multiply numbers is groups that yield a number closest to the denominator. That way you can keep simplifying to smaller numbers and avoid big calculations. Let me know if anything needs to be clarified. Cheers. Hi, What is the trick to find 1056 is a multiple 33? I tried this path but couldn't find the thread: GMAT Home >> GMAT Test Prep Questions >> Number Systems... Thanks,



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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28 May 2012, 10:46
Hi keiraria, Here is the url where you can find examples using the above mentioned tricks: findthelast2digitsof86325.htmlThanks, Kartik



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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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14 Jun 2012, 14:03
Great thread, This is going to give me some good edge.. Thanks a lot Sriharimurthy




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