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If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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If n = (33)^43 + (43)^33 what is the units digit of n? A. 0 B. 2 C. 4 D. 6 E. 8
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Originally posted by Archit143 on 04 Oct 2012, 00:13.
Last edited by Bunuel on 04 Oct 2012, 04:23, edited 1 time in total.
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Re: If n = (33)^43 [#permalink]
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Archit143 wrote: If n = (33)^43 + (43)^33 what is the units digit of n?
0 2 4 6 8
pls suggest approach for such probs In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1) Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9, 7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7, 1) Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=1 0 = unit digit is 0



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Re: If n = (33)^43 [#permalink]
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Archit143 wrote: If n = (33)^43 + (43)^33 what is the units digit of n?
0 2 4 6 8
pls suggest approach for such probs Hi Archit, Just follow the unit digits: n = (33)^43 + (43)^33 Here the last digits would determine the power. So, n = (33)^43 + (43)^33 ~ 3^43 + 3 ^33 Now last digits of 3^1, 2, 3, 4, 5 = 3, 9, 7, 1, 3. The cycle repeats after every 4 rounds. So n = 3^43 + 3 ^33 = 3^(40+3) + 3^(32+1) = {(3^(4*10)}{3^3} + {(3^(4*8)}{3^1} Now last digits of these terms would be {1}{7} + {1}{3} = 10 Hence the last digit is 0. Hope this helps. Regards, Shouvik
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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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Both terms has 3 at the units place, so the units place cycle around 3,9,7,1 33^43 will have 7 & 43^33 will have 3 7 + 3 = 10, 0 would be in the units place of the resultant Answer = 0 = A
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Re: If n = (33)^43 [#permalink]
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bhavinshah5685 wrote: Archit143 wrote: If n = (33)^43 + (43)^33 what is the units digit of n?
0 2 4 6 8
pls suggest approach for such probs In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1) Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9, 7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7, 1) Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=1 0 = unit digit is 0I can't understand why it has to be a multiple of 4. I mean, 43 can also be expressed as (2)(21)+1 but this would be wrong. Generally in such exercises it has to be a multiple of 4, or I don't get something?



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Chrysopigi89 wrote: bhavinshah5685 wrote: Archit143 wrote: If n = (33)^43 + (43)^33 what is the units digit of n?
0 2 4 6 8
pls suggest approach for such probs In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1) Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9, 7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7, 1) Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=1 0 = unit digit is 0I can't understand why it has to be a multiple of 4. I mean, 43 can also be expressed as (2)(21)+1 but this would be wrong. Generally in such exercises it has to be a multiple of 4, or I don't get something? If n = (33)^43 + (43)^33 what is the units digit of n?A. 0 B. 2 C. 4 D. 6 E. 8 First of all, the units digit of (33)^43 is the same as that of 3^43 and the units digit of (43)^33 is the same as that of 3^33. So, we need to find the units digit of 3^43 + 3^33. Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}: 3^1=3 (the units digit is 3) 3^2=9 (the units digit is 9) 3^3=27 (the units digit is 7) 3^4=81 (the units digit is 1) 3^5=243 (the units digit is 3 again!) ... Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1). Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0. Answer: A.
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Re: If n = (33)^43 [#permalink]
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Bunuel wrote: If n = (33)^43 + (43)^33 what is the units digit of n?
A. 0 B. 2 C. 4 D. 6 E. 8
First of all, the units digit of (33)^43 is the same as that of 3^43 and the units digit of (43)^33 is the same as that of 3^33. So, we need to find the units digit of 3^43 + 3^33.
Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}: 3^1=3 (the units digit is 3) 3^2=9 (the units digit is 9) 3^3=27 (the units digit is 7) 3^4=81 (the units digit is 1) 3^5=243 (the units digit is 3 again!) ...
Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).
Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.
Answer: A. For more on this check theory and problems listed below: Problem Solving: 650+ whatisthetensdigitof127023.htmlifyoudivide7131by5whichremainderdoyouget83350.htmlifnisapositiveintegerwhatistheremainderwhen105067.htmlwhatistheunitsdigitof101015.htmlifnisapositiveintegerwhatistheremainderwhen96262.htmlwhatistheremainderwhen3243isdividedby141050.htmlfindtheonesdigitof141071.htmlm1272970.html700+ when5125isdividedby13theremainderobtainedis130220.htmlm2573474.htmlwhichofthefollowingnumbersisprime168179.htmlwhatistheremainderwhen4386isdividedby134778.htmlwhen5125isdividedby13theremainderobtainedis130220.htmlwhatistheremainderwhen4371743628232isdividedby154889.htmlifn33434333whatistheunitsdigitofn140037.html750+ whatistheunitsdigitof126681.htmlwhatistheremainderof126493.htmlwhatistheremainderwhen323232isdividedby100316.htmlalgebram26145109.htmlwhatistheremainderwhen333222isdividedby156379.htmlwhatistheremainderwhen182210isdividedby99724.htmlData Sufficiency: 600+ ifkisapositiveintegerwhatisthereminderwhen2k126478.html650+ ifxisapositiveintegeristheremainder0when3x109075.htmlif243x463ynwherexandyarepositiveintegers102054.htmlifrsandtareallpositiveintegerswhatisthe136746.html700+ toughandtrickyexponentsandrootsquestions125967.html#p1029239ifxandyarepositiveintegerswhatistheremainderwhen109636.htmlHope this helps.
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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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09 Sep 2014, 02:46
Bunuel, I found mistake in the solution given by you The Cycle here is 9, 7, 1, 3 Not 3, 9, 7 , 1 (as you stated) Multiply 33 x 33 Once, what do we get 1089, Unit digit is 9. So the cycle should start at 9 By that logic we should add 1 + 9 = 10 Answer is still A. But lets tweak it (33)^43 + (43)^34 Bunuel you will do this with your chosen cycle 7+9 = 16, Your answer will be 6 With my chosen Cycle 1+7 = 8. That could be a trap if they play on the starting point of the cycle.
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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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09 Sep 2014, 07:13
honchos wrote: Bunuel,
I found mistake in the solution given by you
The Cycle here is
9, 7, 1, 3 Not 3, 9, 7 , 1 (as you stated)
Multiply 33 x 33 Once, what do we get 1089, Unit digit is 9. So the cycle should start at 9
By that logic we should add
1 + 9 = 10 Answer is still A.
But lets tweak it
(33)^43 + (43)^34
Bunuel you will do this with your chosen cycle 7+9 = 16, Your answer will be 6
With my chosen Cycle 1+7 = 8.
That could be a trap if they play on the starting point of the cycle. Please reread the solution and follow the links given above. You should start with power of 1: 3^1=3 (the units digit is 3) 3^2=9 (the units digit is 9) 3^3=27 (the units digit is 7) 3^4=81 (the units digit is 1) 3^5=243 (the units digit is 3 again!) ...
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If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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Based on the Number Theory > Last digit of a power (in the GMATclub forum, I cannot link to the post) Step 1: Find the last digit of 33^43find the cyclicity of 33 33^1= _3 33^2= _9 33^3= _7 33^4= _1 33^5= _3 > 33 has the cyclicity of 4 Divide the power by the cyclicity: 43/4 > remainder is 3 (refer to the 3rd position in the cyclicity) > the last digit of 33^43 is the same as that of 33^3 > 7 is the last digitStep 2: Similarly, find the last digit of 43^3343 also has the cyclicity of 4 (ends with 3) Divide 33 by 4 > remainder is 1 > refer to the 1st position in the cyclicity > last digit is 3 Step 3: 7+3 = 10 > the last digit is 0 Answer A



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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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10 May 2015, 01:50
Bunuel wrote: Thus: The units digit of 3^43 is the same as the units digit of 3^3, so 7 (43 divided by the cyclicity of 4 gives the remainder of 3). The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).
Therefore the units digit of (33)^43 + (43)^33 is 7 + 3 = 0.
Answer: A. I do not understand this part with remainder and cyclicity, why do i need to divide 43 by 4 to get a remainder of one which is then the units digit??



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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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Archit143 wrote: If n = (33)^43 + (43)^33 what is the units digit of n?
A. 0 B. 2 C. 4 D. 6 E. 8 the unit digit of 3^1=3 3^5=3 3^9=3 3^33=3 therefore, 3^33+3^33=6 the correct answer is D



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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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Archit143 wrote: If n = (33)^43 + (43)^33 what is the units digit of n?
A. 0 B. 2 C. 4 D. 6 E. 8 This is a units digit pattern question. The first thing to recognize is that in units digit pattern questions we only care about the units digit place value. Thus, we can rewrite the problem as: (3)^43 + (3)^33 We now need to determine the units digit of (3)^43 + (3)^33. Let's determine the pattern of units digits that we get when a base of 3 is raised to consecutive exponents. 3^1 = 3 3^2 = 9 3^3 = 27 (units digit of 7) 3^4 = 81 (units digit of 1) 3^5 = 243 (units digit of 3) Notice at 3^5, the pattern has started over: 3^6 = units digit of 9 3^7 = units digit of 7 3^8 = units digit of 1 So we can safely say that the base of 3 gives us a units digit pattern of 3, 9, 7, 1, 3, 9, 7, 1, …) that repeats every four exponents. Also notice that every time 3 is raised to an exponent that is a multiple of 4, we are left with a units digit of 1. This is very powerful information, which we can use to solve the problem. Let’s start with the units digit of (3)^43. An easy way to determine the units digit of (3)^43, is to find the closest multiple of 4 to 43, and that is 44. Thus we know: 3^44 = units digit of 1 So we can move back one exponent in our pattern and we get: 3^43 = units digit of 7 Let’s now determine the units digit of (3)^33. We already know that the pattern of units digits for powers of 3 will be 3, 9, 7, 1, 3, 9, 7, 1, … An easy way to determine the units digit of (3)^33 is to find the closest multiple of 4 to 33, and that is 32. Thus we know: 3^32 = units digit of 1 So we can move up one exponent in our pattern and we get: 3^33 = units digit of 3 The last step is to add the two units digits together so we have: 7 + 3 = 10, which has a units digit of zero) Answer is A.
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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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Units digit of a product will ALWAYS depend on the units digits of the numbers being multiplied.Just focus on the units digit. Units digit of n=(3)^43+(3)^33 The cyclicity of 3 is 4. 3^1=3 3^2=9 3^3=27 (Units digit 7) 3^4=81 (Units digit 1) After this the digits will start repeating (3,9,7,1,3,9,7,1, and so on) A the cyclicity is 4 (here), write the powers as multiple of 4 and look for the remainder (if any). 43=4*10+ 3Therefore the units digit of (3)^43= 7 33=4*8+1 Therefore the units digit of (3)^33= 3 The units digit of n= 7+3=10 (Just consider the last digit i.e. the unit's digit) The answer is 0.
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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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29 Dec 2017, 15:02
How is this only a 600 to 700 level question? Manhattan GMAT has it listed as "Devilish" under the difficulty.



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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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30 Dec 2017, 08:18
bhavinshah5685 wrote: Archit143 wrote: If n = (33)^43 + (43)^33 what is the units digit of n?
0 2 4 6 8
pls suggest approach for such probs In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1) Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9, 7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7, 1) Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=1 0 = unit digit is 0Hello bhavinshah5685 if exponent 33=4(8)+1 and unit digit is 3^1 than why did you highligh 1 in red instead of 3 ? thanks!



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Re: If n = (33)^43 + (43)^33 what is the units digit of n? [#permalink]
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31 Dec 2017, 11:06
dave13 wrote: bhavinshah5685 wrote: Archit143 wrote: If n = (33)^43 + (43)^33 what is the units digit of n?
0 2 4 6 8
pls suggest approach for such probs In such type of exponents problem, first find the cyclicity of unit digit of 3. which is (3,9,7,1) Now convert exponent value in to multiple of 4. here 43=4(10)+3 so take 3^3 as unit value of first term which is 27 ===> unit digit is 7 (3,9, 7,1) Now 33=4(8)+1,so take 3^1 as unit value of second term which is 3^1====> unit digit is 3 (3,9,7, 1) Now unit digit of n = unit digit of first term + unit digit of 2nd term = 7+3=1 0 = unit digit is 0Hello bhavinshah5685 if exponent 33=4(8)+1 and unit digit is 3^1 than why did you highligh 1 in red instead of 3 ? thanks! He must have done it by mistake and highlighted 1 after getting the remainder as 1. By the way, his approach and answer are absolutely correct. Regards, Saquib eGMATQuant Expert
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