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\(R of (201*202*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/100\)
\(= R of (201*101*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/50\)
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!]
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
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 \(-6+76 = 70\)
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!!
I have solved these questions here (Two similar topics are merged-Moderator). 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.
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, 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.
It seems there's a shortcut for the second problem: 1.) Forget about 201 and 246 and simplify it to 1*2*3*4*4*3*2*1, since 246 is the same as (250-4) giving 4, not 6. 1*2*3*4*4*3*2*1=2^2*3^2*4^2=574
2.)Now, take 74 out of 574 and calculate a square of 74. It doesn't matter what it is but it ends with 76
Last edited by felixjkz on 10 Dec 2012, 00:14, edited 1 time in total.
Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
29 Apr 2013, 02:47
1
This post received KUDOS
Expert's post
sdas wrote:
Hi Karishma, many thanks for your response. I am a fan of your blogs, do see my Quant Consolidated notes, Carcass's efforts of your blogs are best in there...
My confusion is how do we get to remainder 6 and not -19? 6/25 if this has remainder 6 then why should 24/25 have remainder of -1 (is the formula not same (x-n)/n The negative part I understand completely, your blogs are too good
When the divisor is 25, a remainder of 6 is the same thing as a remainder of -19. Since GMAT doesn't give you negative remainders, you will not have -19 in the options. So you mush choose 6. 24/25 has remainder 24 which is also same as remainder of -1. Both are correct but again, GMAT will only give you 24 in the options. Think of it: when you have 24 balls and you must distribute them equally among 25 kids, you can say that you have 24 balls remaining and you gave each kid 0 (the quotient) ball. Or you can say that you have -1 ball remaining (i.e. you gave one extra ball from your side) and each kid got 1 (the quotient) ball. _________________
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? _________________
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