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Re: Find the last two digits [#permalink]
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02 Dec 2010, 11:49
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.
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Re: Numbers 4 [#permalink]
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02 Dec 2010, 11:55
Of course last 2 digits of the first question can be easily obtained: 65*29*37*63*71*87*62 1. 70 2. 30 3. 10 4. 90 We have a 5 (in 65) and a 2 ( in 62) So 65*29*37*63*71*87*62 = 5*13 * 29*37*63*71*87*2 * 31 10 *13 * 29*37*63*71*87*2 * 31 The last digit will be 0. The second last digit will be the last digit of 13*29*37*63*71*87*31 which we get as 9. So last two digits will be 90.
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Re: Find the last two digits [#permalink]
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07 May 2011, 09:19
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). In your operation you are doing division to make the problem simpler specifically, Quote: 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. You cannot divide to simplify when you are setting out to find a remainder. For example what is the remainder of \(14/10\)? 4 what is the remainder of \(7/5\)? 2 Notice that both 14/10 and 7/5 are both equivalent. But the remainders are different



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Re: Find the last two digits [#permalink]
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07 May 2011, 19:25
gprasant wrote: You cannot divide to simplify when you are setting out to find a remainder. For example what is the remainder of \(14/10\)? 4 what is the remainder of \(7/5\)? 2 Notice that both 14/10 and 7/5 are both equivalent. But the remainders are different
Yes. You cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, remember, if you do want to cancel off to make life easier for you, you can do it, provided you remember to multiply it back again. So say, I want to find the remainder when 14 is divided by 10 (remainder 4) and cancel off the common 2 to get 7 divided by 5 giving me a remainder of 2, I can multiply back the 2 I canceled with the remainder to get a remainder of 4. That's a valid technique. e.g. What is the remainder when 85 is divided by 20? It is 5. or I might rephrase it as what is the remainder when 17 is divided by 4 (I cancel off 5 from the numerator and the denominator). The remainder in this case is 1. I multiply the 5 back to 1. I get the remainder as 5!
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Re: Find the last two digits [#permalink]
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27 Apr 2013, 12:33
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). Can someone explain the highlighted part. Why R of (xn)/n = R of (625)/25 = +6?
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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
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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 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, 11:55
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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 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 2 digits of 65*29*37*63*71*87*62
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