Mahi111 Posted December 19, 2006 Share Posted December 19, 2006 Can someone please tell me how to find of the unit digit of the following. Find out the unit digit of 4^27 * 5^27 * 3^27? Not only the above one,how to find unit digit of any value to any power? Please some one explain me this concept. Thx. Quote Link to comment Share on other sites More sharing options...
moe Posted December 19, 2006 Share Posted December 19, 2006 5, 15, 25.... to any power, always unit digit=5, and same goes for 6 others follow easy patterns: the easiest are 4 and 9: 4, 6, 4, 6.... and 9, 1... 2^1=2 2^2=4 2^3=8 2^4=16 2^5=32 (back to the start, unit = 2) for 8 it's simmilar to 2: 8, 4, 2, 6 3 and 7 are simmilar too: 3, 9, 7, 1 and 7, 9, 3, 1 helps?![dance] [banana] [dance] Quote Link to comment Share on other sites More sharing options...
moe Posted December 19, 2006 Share Posted December 19, 2006 and to find out the unit digit of 4^27 * 5^27 * 3^27... 4^27 will have unit = 4 and 5^27, unit = 5 ==> 4^27 * 5^27, unit = 0 ==> 4^27 * 5^27 * 3^27, unit = 0 is that the answ? Quote Link to comment Share on other sites More sharing options...
Mahi111 Posted December 19, 2006 Author Share Posted December 19, 2006 Thank you moe,Yes.u r answer is correct. but I did not understand this part and anything to the power of 2, mean that first I should check where it is repeating,then, that reaping one is the unit digit or else ,I strucked up here. 2^1=2 2^2=4 2^3=8 2^4=16 2^5=32 (back to the start, unit = 2) for 8 it's simmilar to 2: 8, 4, 2, 6 3 and 7 are simmilar too: 3, 9, 7, 1 and 7, 9, 3, 1 I think I have not understand the concept perfectly,can you please help me by saying bit clearly. Quote Link to comment Share on other sites More sharing options...
moe Posted December 19, 2006 Share Posted December 19, 2006 ok... lets see if i explain myself any better...:blush: 2^1=2 2^2=4 2^3=8 2^4=16 2^5=32 ==> same unit digit as 2 to the power of 1 (2^1=2) 2^6=64 ==> same unit digit as 2 to the power of 2 (2^2=4) . .. ... units digit can be grouped into a sequence of 4 numbers that repeat themselves 2/4/8/6==> 2 will be the unit number for 2 to the power of 1, 5, 9, 13... and, in the same way, 4 will be the unit number for 2 to the power of 2, 6, 10, 14... eg: what's the unit digit of 2^16? 16/4=4 no remainder, therefore, the units digit will be the last one in the series=6 (and this same sequence applies to the powers of numbers 12, 22, 32... 32^3 will have a unit digit of 8, the same as 22^3 and 2^3) numbers 3, 7 and 8 also have sequences of 4 numbers: for nº3: 3/9/7/1 eg:3^4 will have a unit digit of 1 (and 13^4 as well), 3^5 will have unit digit 3... the sequences of 7 (7, 9, 3, 1) and 8 (8, 4, 2, 6) follow the same rules Quote Link to comment Share on other sites More sharing options...
sruthi07 Posted December 19, 2006 Share Posted December 19, 2006 explanation moe, and time saving too!! thank you! i had a diff approach, let me know whther im right viz; a ^b * c^d * e ^f ; unit digit of this one will be a^1* b^1* c^1...??? so in the problem above 5*4*3 has a unit digit = 0 Quote Link to comment Share on other sites More sharing options...
moe Posted December 19, 2006 Share Posted December 19, 2006 i'm sorry, but i'm not sure i get the method... if i understand well, in a shorter example, like 4^2*3^7, u would do 4*2*3*7? that wouldn't work on this new example... Quote Link to comment Share on other sites More sharing options...
gre_girl Posted December 20, 2006 Share Posted December 20, 2006 hi moe sory but i cant understand this concept.can u explain again thanks in advance regards Quote Link to comment Share on other sites More sharing options...
whaw Posted July 24, 2008 Share Posted July 24, 2008 helo mahi111, u can solv for the units digit by getting the patern...here's how 4^1=4 4^2=6 4^3=4 4^4=6 So 27/2=13 remainder 1 therefore units digit is 4. 5^1=5 5^2=5 5^3=5...therefore the units digit is 5...and so on.. Quote Link to comment Share on other sites More sharing options...
taminkeu Posted July 24, 2008 Share Posted July 24, 2008 4^27 * 5^27 * 3^27 = (4*5*3)^27 = 60^27 unit of 60^1 = 0 unit of 60^2 = 0 unit of 60^3 = 0 . . . . unit of 60^27= 0 Quote Link to comment Share on other sites More sharing options...
mystic87 Posted July 24, 2008 Share Posted July 24, 2008 choices, choices, choices... these are all great by the way. really makes you think about the properties of numbers. Quote Link to comment Share on other sites More sharing options...
kobra Posted May 12, 2010 Share Posted May 12, 2010 1. 4^2 * 3^7 unit digit = (4*3) = 12 =2 2. 4^2 unit digit = 6 3^7 unit digit = 7 6*7 = 42 unit dig =2 Quote Link to comment Share on other sites More sharing options...
kobra Posted May 16, 2010 Share Posted May 16, 2010 isnt the unit digit of 3^27 = 7 3 power 1 = 3 2 = 9 3 = 27 4 = 81 3,9,7,1 27/4 = 6 R3 3rd down is 7..... still it would be irrelevant since the unit of the first two terms multiplied end in 0...... Quote Link to comment Share on other sites More sharing options...
Sonal Bhardwaj Posted August 24, 2013 Share Posted August 24, 2013 Find out the unit digit of 4^27 * 5^27 * 3^27? Hi. I have worked out the solution for your question. In the question, we have 4,5,3 as Base. First step is to find out the cyclicity of these numbers. For 3: Base ^ Exponent = Unit Value 3^1 = 3 3^2 = 9 (3x3) 3^3 = 7 (3x9) 3^4 = 1 (3x7) 3^5 = 3 (3x1) and so on. Here, you will observe that the pattern for the unit's digits will repeat after 1, i.e. after counting a total digits of 4. Hence, the cyclicity of 3 =4. For 4: Base ^ Exponent = Unit Value 4^1 = 4 4^2 = 6 (4x4) 4^3 = 4 (4x6) 4^4 = 6 (4x4) and so on. Here, you will observe that the pattern for the unit's digits will repeat after 6, i.e. after counting a total digits of 2. Hence, the cyclicity of 4 =2. For 5: Base ^ Exponent = Unit Value 5^1 = 5 5^2 = 5 (5x5) In the case of 5, each time 5 will be multiplied by 5, unit's place will always have 5. The cyclic value is 1. 5^n will always have 5 in the unit's place. Coming back to the ques, taking the first term, we have: 4^27 We are the find a number less than 27 which must be a fully divisible by 4's cyclic value, i.e. 2. The nearest value is 26. 4^27 = 4^26.4 Here we have to be very careful with the digits. In the cyclic table of 4, the cycle repeats itself after 6. Thus, 6 will be unit's digit for 4^26 as 26%2=0. Substituting 6 in place of 4^26, we have, 4^27 = 4^26.4 = 6.4 = 24 Again, the unit's place for 24 is 4. Set aside this value. Similarly, 5^27 = 5^25.5^2 = 5.5^2 = 125 Again, the unit's place for 25 is 5. Set aside this value. Similarly, 3^27 = 3^24.3^3 In the cyclic table of 3, the cycle repeats itself after 1. Thus, 1 will be unit's digit for 3^24 as 24%4=0. Substituting 1 in place of 3^24, we have, => 1.3^3 = 27 Again, the unit's digit for 27 is 7. Set aside this value too. Combining the 3 values, we have, 4,5 and 7. 4x5x7 will give 140 which will have the unit's place = 0. I hope you will understand this. This is the fastest way to solving such questions. All the best !!!!!! ================================================================================ Can someone please tell me how to find of the unit digit of the following. Find out the unit digit of 4^27 * 5^27 * 3^27? Not only the above one,how to find unit digit of any value to any power? Please some one explain me this concept. Thx. Quote Link to comment Share on other sites More sharing options...
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