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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
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27 Apr 2013, 23:30
virtualanimosity wrote: 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 Q1. For these type of questions, always try to relate the numbers to a common divider. Here it will be 10 and hence we can minimize the products are as follows. 5*1*3*3*1*3*2 = 10*27 = 270 = the final digit will be 0. But to get the tenth digit, please add the remaining two = (2+7). In the end, the answer will be 90. Q2. same approach. 1*2*3*4*4*3*2*1 = 24*24 = 576, so tenth digit will be (5+7) & again (1+2) = 3. Finally the last two digit of the expresiion the bracket will be 36. To find the last two digits after applying the square function is to follow the same procedure. Take the last digit, which is 6. Hence, 6*6 = 36



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Re: Find the last two digits [#permalink]
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27 Apr 2013, 23:48
sdas wrote: Can someone explain the highlighted part. Why R of (xn)/n = R of (625)/25 = +6?
I am not sure from where you are reading this and what this means but I think you are confused with the concept of negative remainders. It is discussed here: http://www.veritasprep.com/blog/2011/05 ... emainders/When the remainder of x/n is 6, there is no confusion. But when the remainder of x/n comes out to be 6, we need to adjust the remainder (make it positive) by making it n  6 (why has been discussed in the post) Also, when talking about divisibility and remainders, you don't usually take negative integers into account. Divisibility is a positive integer concept. You can divided 6 balls among 3 kids but not 6 balls among 3 kids. So (625) = 19 divided by 25 doesn't really make sense. If instead it were (25n  19) divided by 25, then the remainder will be 19 since 25n is divisible by 25. A remainder of 19 is same as a remainder of 6.
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
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28 Apr 2013, 00:54
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 (xn)/n The negative part I understand completely, your blogs are too good
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
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29 Apr 2013, 03:47
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 (xn)/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.
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
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29 Apr 2013, 06:52
Thanks Karishma, for clearing my doubt.
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
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29 Apr 2013, 11:55
virtualanimosity wrote: 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 Q1. 65*29*37*63*71*87*62 1. 70 2. 30 3. 10 4. 90 Finding the last two digits of an expression Thru the Application of REMAINDER THEOREM
Remainder of the above expression when divided by 100 will give the answer of this question \(\frac{65*29*37*63*71*87*62}{100}\) \(\frac{13*29*37*63*71*87*62}{20}\) > on dividing by 5 \(\frac{13*9*17*3*11*7*2}{20}\) > taking remainder when each number divided by 20 \(\frac{117*51*77*2}{20}\) \(\frac{17*11*17*2}{20}\) > taking remainder when each number divided by 20 \(\frac{289*22}{20}\) \(\frac{9*2}{20}\) > taking remainder when each number divided by 20 \(\frac{18}{20}\) So 18 is the remainder However since initially we divided the numerator and denominator by 5, now we need to multiply 18 by 5 So last two digits = 90 Thru the Application of NEGATIVE REMAINDERS
\(\frac{13*29*37*63*71*87*62}{20}\) \(\frac{(7)*9*(3)*3*(9)*7*2}{20}\) > taking remainder when each number divided by 20 \(\frac{21*27*(9)*14}{20}\) \(\frac{1*7*(9)*(6)}{20}\) > taking remainder when each number divided by 20 \(\frac{7*54}{20}\) \(\frac{(13)*(6)}{20}\) > taking remainder when each number divided by 20 \(\frac{78}{20}\) \(\frac{18}{20}\) > taking remainder when each number divided by 20 18*5 = 90 NOTE : If we are asked to find last 3 digits of the expression, we will obtain individual remainders when divided by 1000 Although Indian CAT is fond of such problems, i have never seen those in my any GMAT practice materialRegards, Narenn
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Re: Find the last two digits [#permalink]
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25 Jun 2013, 08:44



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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
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25 Jun 2013, 09:49
riyazv2 wrote: virtualanimosity wrote: 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 Q1. For these type of questions, always try to relate the numbers to a common divider. Here it will be 10 and hence we can minimize the products are as follows. 5*1*3*3*1*3*2 = 10*27 = 270 = the final digit will be 0. But to get the tenth digit, please add the remaining two = (2+7). In the end, the answer will be 90. Q2. same approach. 1*2*3*4*4*3*2*1 = 24*24 = 576, so tenth digit will be (5+7) & again (1+2) = 3. Finally the last two digit of the expresiion the bracket will be 36. To find the last two digits after applying the square function is to follow the same procedure. Take the last digit, which is 6. Hence, 6*6 = 36 How would you find the last two digit of 33*33*33*33 using the method mentioned here.



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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
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25 Jun 2013, 22:34



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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
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27 Jun 2013, 09:21
prateekbhatt wrote: koolgmat wrote: riyazv2 wrote: Q1.
For these type of questions, always try to relate the numbers to a common divider. Here it will be 10 and hence we can minimize the products are as follows.
5*1*3*3*1*3*2 = 10*27 = 270 = the final digit will be 0. But to get the tenth digit, please add the remaining two = (2+7). In the end, the answer will be 90.
Q2.
same approach. 1*2*3*4*4*3*2*1 = 24*24 = 576, so tenth digit will be (5+7) & again (1+2) = 3. Finally the last two digit of the expresiion the bracket will be 36.
To find the last two digits after applying the square function is to follow the same procedure. Take the last digit, which is 6. Hence, 6*6 = 36 How would you find the last two digit of 33*33*33*33 using the method mentioned here. 33*33*33*33 /100 33*33*33*33 /5 > Divide the denominator by 20. 2*2*2*2/5= 16/5 = 1 Now lets multiply 1 with 20 (since we divided it with 20) = 20 If i apply this logic to find last two digit of 77*77*77*77 then 77*77*77*77/100 77*77*77*77/5 > Dividing the denominator by 20. 2*2*2*2 /5 = 16/5 = 1 We multiply 1 with 20 ( since we divided it by 20) = 20 This gives answer as 21 and not 41. Please help explain the logic when we multiply and when we do not.



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Re: Find the last two digits [#permalink]
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20 Sep 2013, 20:25
sriharimurthy wrote: 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). This is a great method. However i am not able to understand why R of 246/50 = 4? Please help to explain.
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Re: Find the last two digits [#permalink]
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21 Sep 2013, 06:25
ankur1901 wrote: This is a great method. However i am not able to understand why R of 246/50 = 4? Please help to explain. No, 4 is the Negative Remainder and not actual remainder. We can get actual remainder by adding divisor in the negative remainder i.e. 4+50 = 46 Refer my this post, if you still have any query about remainder theorem : findthelast2digitsof8632560.html#p1218383
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
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21 Sep 2013, 12:08
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virtualanimosity wrote: 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 in fact, it is also not that time consuming to multiply all numbers since we do not have to calculate exact and entire digits off the resulting number. All we need is last two digits, so try multiply them all, it is still workable when you cannot think of any shortcuts when in a hurry




Re: Find the last 2 digits of 65*29*37*63*71*87*62
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