Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

How do you get this? As in is this a formula??? Please guide.. _________________

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|

I am sure its not a GMAT question, even after using techniques it takes more than 2-3 min.. Here we go: (201*202*203*204*246*247*248*249)^2

we can make pairs within the numbers inside the brackets such as (200+1)(200+49) which gives us last 2 digits = 49 (200+2)(200+48) which gives us last 2 digits = 96 (200+3)(200+47) which gives us last 2 digits = 141 (200+4)(200+46) which gives us last 2 digits = 184

No again the last two dogits can be paired as 49*141 = (100-41)(100+41) or (100^2 - 41^2) which gives us last 2 digits = 19 and 96*184 = (140-44)(140+44) or (100^2 - 44^2) which gives us last 2 digits = 64

Now, 19*64 gives us last 2 digits of 16 and square of 16 will be 56. Phewwww!!

Pls tell me I am right, else i am not attempting the first one!!

49=100-51 but not 41. Or it doesn't matter?

Could you also explain how to calculate that (100^2 - 44^2) has 64 as last two digits?

concept to be used for such sums is REMAINDER THEOREM

to get last 2 digits divide by 100

(65*29*37*63*71*87*62)/100=

13*29*37*63*71*87*62)/20= ....dividing by 5 both numerator n deno

Remainder Thm---> -7*9*-3*3*11*7*2/20 ( ie 13/20 gives us remainder -7 or 13;29/20 gives us rem 9....... = -63*-99*14/20 = 63*99*14/20 Remainder Thm--->3*-1*-6/20 = 18/20 that gives us remainder 18.....but 1st step we had divided by 5 therfore multiply by 5 now ie remainder = 18*5=90

but from 1st method it comes to 70 (5*1*3*3*1*3*2)=270 2+7 =9 so is it 90

I am sure its not a GMAT question, even after using techniques it takes more than 2-3 min.. Here we go: (201*202*203*204*246*247*248*249)^2

we can make pairs within the numbers inside the brackets such as (200+1)(200+49) which gives us last 2 digits = 49 (200+2)(200+48) which gives us last 2 digits = 96 (200+3)(200+47) which gives us last 2 digits = 141 (200+4)(200+46) which gives us last 2 digits = 184

No again the last two dogits can be paired as 49*141 = (100-41)(100+41) or (100^2 - 41^2) which gives us last 2 digits = 19 and 96*184 = (140-44)(140+44) or (100^2 - 44^2) which gives us last 2 digits = 64

Now, 19*64 gives us last 2 digits of 16 and square of 16 will be 56. Phewwww!!

Pls tell me I am right, else i am not attempting the first one!!

49=100-51 but not 41. Or it doesn't matter?

Could you also explain how to calculate that (100^2 - 44^2) has 64 as last two digits?

concept to be used for such sums is REMAINDER THEOREM

to get last 2 digits divide by 100

(65*29*37*63*71*87*62)/100=

13*29*37*63*71*87*62)/20= ....dividing by 5 both numerator n deno

Remainder Thm---> -7*9*-3*3*11*7*2/20 ( ie 13/20 gives us remainder -7 or 13;29/20 gives us rem 9....... = -63*-99*14/20 = 63*99*14/20 Remainder Thm--->3*-1*-6/20 = 18/20 that gives us remainder 18.....but 1st step we had divided by 5 therfore multiply by 5 now ie remainder = 18*5=90

but from 1st method it comes to 70 (5*1*3*3*1*3*2)=270 2+7 =9 so is it 90

Re: Find the last two digits [#permalink]
20 Apr 2010, 20:40

1

This post received KUDOS

Hussain15 wrote:

sriharimurthy wrote:

Now, \(R of (13*29*37*63*71*87*31)/10 = R of [3*(-1)*(-3)*3*1*(-3)*1]/10 = R of -81/10 = -1\)

I have read the post & read the tips & tricks on remainders too but could not understand how srihari is getting \(-1, -3 & -3\) in above equation?

Can anyone help me out?

\(-1, -3 & -3\) came after dividing \(29,37 & 87\) with \(10\), as -ve remainder. this is approach to solve such problem. when -ve value came 10 is added because remainder is always +ve.

if this approach is troubling you, you can still solve as follows by doing division normal way:

\(R of (13*29*37*63*71*87*31)/10 = R of [3*9*7*3*1*7*1]/10\)

or \(R of [3*9*7*3*1*7*1]/10 = R of [27*21*7]/10\)

Re: Find the last two digits [#permalink]
21 Apr 2010, 06:12

sriharimurthy wrote:

Hi, there is a very quick way to solve these questions:

Since we want to find the last two digits, we have to find the remainder when divided by 100. (in decimal format)Therefore,

\(R of (65*29*37*63*71*87*62)/100 = R of (13*29*37*63*71*87*31)/10\) Note: Divided 65 by 5 and 62 by 2.

Now, \(R of (13*29*37*63*71*87*31)/10 = R of [3*(-1)*(-3)*3*1*(-3)*1]/10 = R of -81/10 = -1\)

Since remainder is coming negative, we have to add it to 10.

Thus remainder is in fact 9. In decimal format, it is expressed as 0.9 or (9/10)

Thus the remainder when 65*29*37*63*71*87*62 is divided by 100 in decimal format is 0.9.

The last two digits will therefore be 0.9*100 = 90.

Thus answer is (4).

Also, can anyone explain the red bold portion highlighted above?? I have read the post of tips & tricks on remainders but couldn't understand this sentence. _________________

I just couldn't understand the \(-ve\) remainders, I was aware of only positive values like \(Rof 29/10=9\).

Can you elaborate that how do you get \(Rof 29/10=-1\) ?

you are correct, there is no concept of \(-ve\) remainder as such. this is just a way how to get remainder specially if divisor is big and numbers are just short of the divisor. in above example it is easy to work with the remainder concept i.e. the concept you have mentioned but have a look on the following problem:

\(\frac{73*75*74}{76}\) if we need to find the remainder of this expression then normal remainder concept will not work, because for each number the remainder is same. in such cases we need to find some other way to work with and that's where \(-ve\) remainder concept came into picture.

ROF of above expression will be \(\frac{-3*-1*-2}{76} = Rof \frac{-6}{76}\) \(-ve\) values we get just counting number required to make remainder zero. e.g. -3 came since we need 3 more to make it 76. similarly others. since \(-ve\) can not be remainder so the actual remainder of the expression will be \(-6+76 = 70\)

Re: Find the last two digits [#permalink]
21 Apr 2010, 07:42

1

This post received KUDOS

Hussain15 wrote:

Also, can anyone explain the red bold portion highlighted above?? I have read the post of tips & tricks on remainders but couldn't understand this sentence.

let us take a simple example, if want to know the unit digit of 9*8 what i will do is just multiply and see the unit digit one other way around to find the unit digit is just divide the expression with 10 and see the reminder, and that will nothing but the unit digit of the expression.

so remainder of the expression \(\frac{9*8}{10} = 2\) and that is what the unit digit of the expression..

based on similar logic we can get last two digits of expression by dividing 100.. you can extend this login up to any number e.g. to get last 3 digits of the any expression, just divide it by 1000 and count the remainder.

hope this will help.

Last edited by einstein10 on 21 Apr 2010, 17:09, edited 1 time in total.

Great explanation again. Got it now. einstein10 you can write the theory of remainderbility.

So if I quote it correctly, on the basis of this concept, I can also find the last two digits of \(44^2\)

Solution:

Last two digits of \(44^2=Rof (44*44)/100\)

\(Rof (44*44)/100=Rof (22*22)/25\)

\(Rof (22*22)/25= Rof (-3*-3)/25=Rof 9/25=9\)

As the remainder 9 is for 25 , so for 100 the remainder will be 36

& I think I got it correct. \(44*44=1936\)

now you are expert. trick here is when u divide 4 both numerator and denominator, you remember to multiply back to get the actual remainder ... one complement you apply concepts very fast..great.

Re: Find the last two digits [#permalink]
22 Apr 2010, 21:06

einstein10 wrote:

Hussain15 wrote:

Also, can anyone explain the red bold portion highlighted above?? I have read the post of tips & tricks on remainders but couldn't understand this sentence.

let us take a simple example, if want to know the unit digit of 9*8 what i will do is just multiply and see the unit digit one other way around to find the unit digit is just divide the expression with 10 and see the reminder, and that will nothing but the unit digit of the expression.

so remainder of the expression \(\frac{9*8}{10} = 2\) and that is what the unit digit of the expression..

based on similar logic we can get last two digits of expression by dividing 100.. you can extend this login up to any number e.g. to get last 3 digits of the any expression, just divide it by 1000 and count the remainder.

hope this will help.

i was also struggling with these problems..very helpful of you.. thanks

Re: Find the last two digits [#permalink]
02 Aug 2010, 19:27

sriharimurthy wrote:

Hussain15 wrote:

Find the last 2 digits Q1. 65*29*37*63*71*87*62 1. 70 2. 30 3. 10 4. 90

Hi, there is a very quick way to solve these questions:

Since we want to find the last two digits, we have to find the remainder when divided by 100. (in decimal format)

Therefore,

\(R of (65*29*37*63*71*87*62)/100 = R of (13*29*37*63*71*87*31)/10\) Note: Divided 65 by 5 and 62 by 2.

Now, \(R of (13*29*37*63*71*87*31)/10 = R of [3*(-1)*(-3)*3*1*(-3)*1]/10 = R of -81/10 = -1\)

Since remainder is coming negative, we have to add it to 10.

Thus remainder is in fact 9. In decimal format, it is expressed as 0.9 or (9/10)

Thus the remainder when 65*29*37*63*71*87*62 is divided by 100 in decimal format is 0.9.

The last two digits will therefore be 0.9*100 = 90.

Thus answer is (4).

Can you please explain how you got -1 below? I can see 3 as 13-10, but how did 29 become a -1? Now, \(R of (13*29*37*63*71*87*31)/10 = R of [3*(-1)*(-3)*3*1*(-3)*1]/10 = R of -81/10 = -1\)

Re: Find the last two digits [#permalink]
01 Dec 2010, 19:52

Hi Guys,

I'm a new member, studying to take the GMAT in the fall. Math is definitely my weak point, and I seem to be struggling with some of the concepts related to remainders, especially this problem. I read the compilation of tips, and those actually make sense. However, it seems to break down in the solution below.

Quote:

\(R of (201*202*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/100\)

In the transition from above to below, I notice the second number '202' is reduce to 101. This seems to be linked to factoring the denominator from 100 to 50, but I can't seem to draw the connection.

Quote:

\(= R of (201*101*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/50\)

Oddly enough, most of the solution after this point makes sense. If anyone is still surfing this thread, any help would be great.

Re: Find the last two digits [#permalink]
02 Dec 2010, 10:49

Expert's post

atherius wrote:

Hi Guys,

I'm a new member, studying to take the GMAT in the fall. Math is definitely my weak point, and I seem to be struggling with some of the concepts related to remainders, especially this problem. I read the compilation of tips, and those actually make sense. However, it seems to break down in the solution below.

Quote:

\(R of (201*202*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/100\)

In the transition from above to below, I notice the second number '202' is reduce to 101. This seems to be linked to factoring the denominator from 100 to 50, but I can't seem to draw the connection.

Quote:

\(= R of (201*101*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/50\)

Oddly enough, most of the solution after this point makes sense. If anyone is still surfing this thread, any help would be great.

202/100 = 101/50 (take 2 common in both terms and cancel it out) This is exactly what has been done above to bring it to lowest terms. It is easier to work with fractions when they are in lowest terms. Here, it makes a lot of sense to have 50 in the denominator since we have 246, 247 etc which can be written as (50 - 4), (50 - 3) etc later. (I think you got that part.) But, when we cancel out common factors in remainders, we need to multiply the remainder obtained with the factors that we canceled at the end. e.g. In the explanation, remainder obtained was 19 when the denominator is 25. But actually denominator was 100 and we canceled out two 2s. So the actual remainder will be 19*2*2 = 76. _________________

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...