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Find the last 2 digits of 65*29*37*63*71*87*62
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04 Nov 2009, 00:14
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Find the last 2 digits Q1. 65*29*37*63*71*87*62 1. 70 2. 30 3. 10 4. 90
Q2. (201*202*203*204*246*247*248*249)^2 1.36 2.56 3.76 4.16



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Re: Numbers 4
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04 Nov 2009, 01:34
65*29*37*63*71*87*62 = (60+5)*(301)*(403)*(60+3)*(70+1)*(903)*(60+2)
multiplying 5*1*3*3*1*3*2 = 90
last two digits are 90



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Re: Numbers 4
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04 Nov 2009, 02:07



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Re: Numbers 4
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04 Nov 2009, 03:09
Ans to 2nd question should be 56. I am sure its not a GMAT question, even after using techniques it takes more than 23 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 = (10041)(100+41) or (100^2  41^2) which gives us last 2 digits = 19 and 96*184 = (14044)(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!!



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Re: Numbers 4
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04 Nov 2009, 14:59
kalpeshchopada7 wrote: Ans to 2nd question should be 56. I am sure its not a GMAT question, even after using techniques it takes more than 23 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 = (10041)(100+41) or (100^2  41^2) which gives us last 2 digits = 19 and 96*184 = (14044)(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=10051 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?
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Re: Numbers 4
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06 Nov 2009, 04:19
Vyacheslav wrote: kalpeshchopada7 wrote: Ans to 2nd question should be 56. I am sure its not a GMAT question, even after using techniques it takes more than 23 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 = (10041)(100+41) or (100^2  41^2) which gives us last 2 digits = 19 and 96*184 = (14044)(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=10051 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



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Re: Find the last two digits
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15 Nov 2009, 13:28
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).
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Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilationoftipsandtrickstodealwithremainders86714.html#p651942
Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html



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Re: Find the last two digits
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Updated on: 15 Nov 2009, 14:07
Quote: Q2. (201*202*203*204*246*247*248*249)^2 1.36 2.56 3.76 4.16 Similarly for this question, \(R of (201*202*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/100\) \(= R of (201*101*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/50\) Note: I have left denominator as 50 since it will be easier in calculations. \(= R of [(1*1*3*4*(4)*(3)*(2)*(1)]*[(1*2*3*4*(4)*(3)*(2)*(1)]/50\) \(= R of (12*24*24*24)/50 = R of (6*24*24*24)/25 = R of [6*(1)*(1)*(1)]/25 = 6\) Since remainder is coming negative, we add 25 to it. Thus Remainder is 19. In decimal format, it is 19/25 or 0.76 Thus last two digits will be 0.76*100 = 76 [Note: Rather than calculating the decimal value first, it will be faster to combine the last two steps as follows: (19/25)*100 = 19*4 = 76. This is how I did it and it saved me valuable seconds!] Answer should be (3).
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Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilationoftipsandtrickstodealwithremainders86714.html#p651942
Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html
Originally posted by sriharimurthy on 15 Nov 2009, 13:37.
Last edited by sriharimurthy on 15 Nov 2009, 14:07, edited 1 time in total.



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Re: Find the last two digits
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15 Nov 2009, 13:44
In case you are not able to follow my steps completely, view my post on tips and tricks to deal with remainders. Take your time with them and make sure you understand all the points properly (especially points 5 and 6). Then try to follow the steps I've done. Once you understand those concepts thoroughly you should be able to solve these questions really fast. In fact, it took me less than 2 minutes to solve each of the above problems. All the best! Cheers.
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Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilationoftipsandtrickstodealwithremainders86714.html#p651942
Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html



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Re: Numbers 4
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15 Nov 2009, 13:48
Hi guys, I have solved these questions here ( Two similar topics are mergedModerator). It took me less than 2 minutes to solve each of these questions. Kindly have a look at my method and try to understand it. It will really help you solve these problems really fast even if they come on the GMAT. Cheers.
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Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilationoftipsandtrickstodealwithremainders86714.html#p651942
Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html



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Re: Numbers 4
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15 Nov 2009, 14:01



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Re: Numbers 4
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15 Nov 2009, 14:14
Bunuel wrote: Excellent! I was just posting the solutions for these two questions with similar remainder approach but no need for them now.
+1 for each.
Thanks Bunuel... Makes it all the more special coming from you!
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Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilationoftipsandtrickstodealwithremainders86714.html#p651942
Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html



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Re: Numbers 4
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19 Dec 2009, 23:42
Tania, Topics are covered in MGMAT Number properties Chap 10 or sriharimurthy post covering remainders... compilationoftipsandtrickstodealwithremainders86714.html#p651942bit hurried and cryptic until you a find a question for each type to solve and understand... its well worth it in the end. DC



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Re: Find the last two digits
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20 Apr 2010, 09:14
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?
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Re: Find the last two digits
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20 Apr 2010, 20:40
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 \(R of [3*9*7*3*1*7*1]/10 = R of [27*21*7]/10\) or \(R of [7*1*7]/10 = R of [49]/10\) => remainder is 9 please refer for details in following link http://gmatclub.com/forum/tipshowtogetremainders92928.htmlhope this will help



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Re: Numbers 4
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21 Apr 2010, 06:06
Great explanation. Kudos!! 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 \(Rof 29/10=1\) ?
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Re: Find the last two digits
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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.
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Re: Numbers 4
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21 Apr 2010, 07:30
Hussain15 wrote: Great explanation. Kudos!!
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 \(6+76 = 70\) hope this will help.



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Re: Find the last two digits
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Updated on: 21 Apr 2010, 17:09
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.
Originally posted by einstein10 on 21 Apr 2010, 07:42.
Last edited by einstein10 on 21 Apr 2010, 17:09, edited 1 time in total.



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Re: Find the last two digits
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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 &nbs
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