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Re: Compilation of tips and tricks to deal with remainders. [#permalink]

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10 Dec 2009, 16:06

brownybuddy wrote:

Wondering if it is a good way to solve it as follows:

.12 = 12/100 = 3/25 implies that the remainder is 3 or multiple of 3.

Comments please.

I tested it with some numbers. (8/5, 9/4, 57/12, 57/15, 57/20) They are work using the way you described. I don't know how to prove it mathematically though. Does anyone know? This is a smart way if it works for all numbers!

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

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11 Dec 2009, 00:24

wilbase wrote:

sriharimurthy wrote:

Eg. \(R of \frac{(23)*(27)}{25} = R of \frac{(-2)*(2)}{24} = -4.\) Now, since it is negative, we have to add it to 25.\(R = 25 + (-4) = 21\)

Is the "24" on the second part of the equation suppose to be "25"?

Yes. It is supposed to be 25. Thanks for spotting that. I have edited it.

brownybuddy wrote:

Wondering if it is a good way to solve it as follows:

.12 = 12/100 = 3/25 implies that the remainder is 3 or multiple of 3.

Comments please.

Yes. It follows from property number 7.

Since we are asked to find the remainder when 's' is divided by 't' and we are given the resulting number, we can write an equation as follows :

Remainder = (Decimal portion of the resulting number) * (Number we are dividing by)

Remainder = 0.12 * t

R = \(\frac{12}{100}*t\) = \(\frac{3}{25}*t\)

So as you can see, the remainder 'R' must be a multiple of '3' provided 't' is an integer.

Since we know that 't' is an integer, we can safely conclude that 'R' is a multiple of '3'.

Note : In cases of remainder problems, even if 't' is not an integer it can be made into an integer. Eg. Remainder of 6/2.5 will be the same as Remainder of 12/5. _________________

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

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13 Dec 2009, 04:14

s = Numerator t = Denominator q = Quotient (>= 0) r = Remainder

where,

s/t = q + r/t

In 64.12, q = 64, so

r/t = .12 = 12/100 = 3/25

where 25 is the least possible value of 't' and 1603 is the least possible value of 's'. The other possible values of 't' will include 25, 50, 75 and so on (i.e. multiple of 25). The values of 'r' and 's' will also change accordingly.

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

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19 Dec 2009, 02:17

Hi ,

I am new on this forum , and i will tell you that what i have gained reading all the posts in the various sections have been mind blowing .

I wish to study indepth before i sit for my GMAT. I am quite ambitious with the kind of score i desire and i think you guys are the best in terms of detailing the requirements .

Please kindly explain the 3rd to 5th Rule on the remainder lecture ... i cant seem to grasp the rules!!

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

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21 Dec 2009, 03:18

Property 3 says that

Quote:

3) If a number has a remainder of ‘r’, 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 21 is 5, then remainder of 7 (which is a factor of 21) will also be 5.

But if we see 21/5 remainder is 1 7 is a factor of 21 7/5 and the remainder is 2.

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

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21 Dec 2009, 04:56

msunny wrote:

Property 3 says that

Quote:

3) If a number has a remainder of ‘r’, 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 21 is 5, then remainder of 7 (which is a factor of 21) will also be 5.

But if we see 21/5 remainder is 1 7 is a factor of 21 7/5 and the remainder is 2.

Why the contradiction ??

Hi,

To make this example more clear : If any number when divided by 21 leaves a remainder of 5, then that number when divided by any factor of 21 will also leave a remainder of 5 provided the remainder is less than the factor.

Eg. R of 26/21 = 5

Factors of 21 are 3 and 7

Since 7 is greater than 5, R of 26/7 = 5

Since 3 is less than 5, R of 26/3 = R of 5/3 = 2

Hope this makes it clear.

I think I will edit the main post to make this point less confusing.

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

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21 Dec 2009, 05:17

isiadeolumide33 wrote:

Hi ,

I am new on this forum , and i will tell you that what i have gained reading all the posts in the various sections have been mind blowing .

I wish to study indepth before i sit for my GMAT. I am quite ambitious with the kind of score i desire and i think you guys are the best in terms of detailing the requirements .

Please kindly explain the 3rd to 5th Rule on the remainder lecture ... i cant seem to grasp the rules!!

Thank you

Easy

Hi,

3rd Rule : I have explained the 3rd rule in the above post.

4th Rule : The cycle of powers is useful to know because it tells us the only possible values that the units place can hold for any particular number when it is raised to an integer power.

Go through the following example to see the usefulness of this rule :

Quote:

If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ? (1) n = 2 (2) m = 1

The Concept tested here is cycles of powers of 3.

The cycles of powers of 3 are : 3,9,7,1

St I) n = 2. This makes 3^(4*2 +2 + m) = 3^(10+m). we do not know m and hence cannot figure out the unit digit.

St II) m=1 . This makes 3^(4*n +2 + 1). 4n can be 4,8,12,16... 3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7. SUFF Hence B

5th Rule : Again for this rule, the best way to understand it is to work through a couple of questions (numbers-86325.html). Go through my solutions for the two problems in the post I have linked and see how rules 5 and 6 relate to them.

It might take a while for these concepts to get cemented but have a little patience and you will be rewarded.

If you have any specific doubts you would like me to address then please let me know.

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?

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.

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

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31 Jan 2010, 02:56

sriharimurthy wrote:

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 \(n^x\).[/size]

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.

Hi sriharimurthy...... Thanks a lot for this post...its outstanding..... Kudos x 10 for this

I have a few questions on the cycle of powers part... didnt get this too well...

2: 2, 4, 8, 6 → all \(2^{4x}\) will have the same last digit. Why do you show this as \(2^{4x}\)... what is \(4x\) all about.... ? _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Compilation of tips and tricks to deal with remainders. [#permalink]

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31 Jan 2010, 04:42

sriharimurthy wrote:

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 \(R of \frac{(R of \frac{(x-n)}{n})*(R of \frac{(y-n)}{n})}{n}\) 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 : numbers-86325.html]

I have a question here.....

You explained the remainder formula as :

Quote:

size=130]6) \(R of \frac{x*y}{n}\) , can sometimes be easier calculated if we take it as \(R of \frac{(R of \frac{(x-n)}{n})*(R of \frac{(y-n)}{n})}{n}\) [/size] Especially when x and y are both just slightly less than n.

And in you example:

Quote:

Eg. \(R of \frac{(19)*(21)}{25} = R of \frac{(-6)*(-4)}{25} = 24\)

shouldn't this be Eg. \(R of \frac{(19)*(21)}{25} = R of \frac{(R of \frac{(19-25)}{25})*R of \frac{(21-25)}{25}}{25} = R of \frac{(R of \frac{(-6)}{25})*R of \frac{(-4)}{25}}{25}\)????

After this I am upside down _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Which of the following must be a divisor of a? [#permalink]

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02 Apr 2010, 02:29

Great post. Was very useful. Thanks. _________________

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