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Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]
 04 Nov 2009, 01:14
Question Stats:
27% (02:12) correct
72% (01:07) wrong based on 18 sessions
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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65*29*37*63*71*87*62 = (60+5)*(30-1)*(40-3)*(60+3)*(70+1)*(90-3)*(60+2)
multiplying 5*1*3*3*1*3*2 = 90
last two digits are 90
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Ans to 2nd question should be 56. 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!! 
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Hussain15 ..... Please explain how you reached to ans A
may be your answer is correct
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It looks like that it is not GMAT question... Not reasonable to do in less than 2 minutes.
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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 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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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 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
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I guess that mistake of considering 49 = 100-41 instead of 100-51 was a big one and changed the entire answer. I am sorry guys!! Is the answer 76??
For Vyacheslav, 100^2-44^2 = will give us 10000 - 44^2 which ends with 36, hence the subtraction should result in 64.
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guys, thanks for your explanations. 1) I just wanted to ask whether it is a quick approach to determine the last two digits of 44^2. 2) Could you explain the remainder theorem or give link on it? Thanks in advance.
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these are not gmat question.Kindly mention the source
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Find the last two digits [#permalink]
14 Nov 2009, 07:30
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 These questions were posted on Nov 04, 2009 on this forum. I dont know how these can be solved?? Bunuel can you help on this one??
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Re: Find the last two digits [#permalink]
15 Nov 2009, 14:28
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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 = -1Since 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!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942
Word Problems Made Easy!
1) Translating the English to Math : word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : distance-speed-time-word-problems-made-easy-87481.html
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Re: Find the last two digits [#permalink]
15 Nov 2009, 14:37
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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)/50Note: 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 = -6Since 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!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942
Word Problems Made Easy!
1) Translating the English to Math : word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : distance-speed-time-word-problems-made-easy-87481.html
Last edited by sriharimurthy on 15 Nov 2009, 15:07, edited 1 time in total.
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Re: Find the last two digits [#permalink]
15 Nov 2009, 14:44
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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!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942
Word Problems Made Easy!
1) Translating the English to Math : word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : distance-speed-time-word-problems-made-easy-87481.html
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Hi guys, 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. Cheers.
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Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942
Word Problems Made Easy!
1) Translating the English to Math : word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : distance-speed-time-word-problems-made-easy-87481.html
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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!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942
Word Problems Made Easy!
1) Translating the English to Math : word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : distance-speed-time-word-problems-made-easy-87481.html
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Yet another great post.
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I am not able to understand why in the reminder problems, it is divided by 100?
Can someone explain me?
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close
