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Re M2531
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16 Sep 2014, 01:24



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Re: M2531
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04 Oct 2014, 22:00
why cant we start from 3^0 instead of 3^1 ? I did in this way: 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27 then the cycle repeat with 3^4 = 81, 3^5 =243 the pattern here is (1,3,9,7) the answer should be e ?



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05 Oct 2014, 02:16
langtuprovn2007 wrote: why cant we start from 3^0 instead of 3^1 ? I did in this way: 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27 then the cycle repeat with 3^4 = 81, 3^5 =243 the pattern here is (1,3,9,7) the answer should be e ? That's just wrong. If this were the case, then all numbers in power which is a multiple of cyclicity would have the last digit of 1. You should always start with power of 1. For more check UNITS DIGITS, EXPONENTS, REMAINDERS PROBLEMS in our Special Questions Directory. Hope it helps.
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Re: M2531
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05 Oct 2014, 03:09
can you give me some examples plz !



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05 Oct 2014, 09:08



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Re: M2531
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07 Nov 2015, 13:44
My 2 cents: It doesn't matter whether you start your powers at 0 or at 1; all that matters is recognizing the pattern. let mn = 0,1,2,3,...; then let last digit = x = 1,3,9,7,....; Because x repeats in these 4 numbers; we can tell that x = 1 implies mn = 0,4,8,12....; x=3 implies mn = 1,5,9...; x=9 implies mn = 2,6,10,...; x=7, implies mn = 3,7,11,.... The question that becomes, which series gets you to 27? x=1 and x =3 gives evens, so rule out, leaving x=3 and x=7. From here you can manually add 4 to each series and see which one gets to 27. The answer is x=7. Or if you like algebra, find the series formula for each. For x=3; we have mn = 1,5,9...= 4a 3. There is no integer a such that 4a3 = 27. On the other hand, for x=7, we have mn = 3,7,11,... or 4a1. And 4a1=27 yields an integer value of a, meaning that this series will reach 27



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Re: M2531
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21 Apr 2018, 05:18
Hi guys,
when we divide 27 by 3, that yields a remainder of 3 (maximum possible remainder since R < Divisor)
Now when:
3^1 = 3 R=0 3^2 = 9 R=1 3^3 = 27 R=2 3^4 = 81 R=3
Now since the remainder was 3, I've associated to the last digit of 1.
Can anyone please tell me where I've went wrong in my thought proccess?
Thanks



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Re: M2531
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21 Apr 2018, 05:31
Miracles86 wrote: Hi guys,
when we divide 27 by 3, that yields a remainder of 3 (maximum possible remainder since R < Divisor)
Now when:
3^1 = 3 R=0 3^2 = 9 R=1 3^3 = 27 R=2 3^4 = 81 R=3
Now since the remainder was 3, I've associated to the last digit of 1.
Can anyone please tell me where I've went wrong in my thought proccess?
Thanks You should divide the power, which is 27, by cyclicity number, which is 4 to find the remainder: 27 divided by 4 gives the remainder of 3. Hence, the units digit of 3^27 is the same as the units digit of 3^ 3, which is 7.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: M2531
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21 Apr 2018, 05:35
Got it! I was confusing this exercise with those when u need to find the digit of a non terminating repeating decimal. On those, the value of the remainder gets you straight to the answer...
Thanks a lot!



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Re: M2531
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19 Jul 2018, 06:16
First you should solve for the index i.e. 3 raise to 3 = 27
den 3 raise to 27
so cycle of 4 is. 3 , 9, 7, 1.
27 comes 3rd in this cycle... so answer will be 7
Hope its clear now.










