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Guys I have a querie: 3 has a cycle of four: 3^0=1 3^1=3 3^2=9 3^3=7 3^4=1 So the cycle is 1,3,9,7,,,, Now we need 3^27.27 is 4*6+3 which is 6 full cycles of 4.Remainder is 3.So the 3rd. no. in the cycle is the answer which is 7. HELP!!!! _________________
Guys I have a querie: 3 has a cycle of four: 3^0=1 3^1=3 3^2=9 3^3=7 3^4=1 So the cycle is 1,3,9,7,,,, Now we need 3^27.27 is 4*6+3 which is 6 full cycles of 4.Remainder is 3.So the 3rd. no. in the cycle is the answer which is 7. HELP!!!!
Are you worried that you have got the rigth by using a perfect method!! Why do you need help? Yours is a perfect solution. _________________
Here the given number is \((xyz)^n\) z is the last digit of the base. n is the index
To find out the last digit in \((xyz)^n\), the following steps are to be followed. Divide the index (n) by 4, then
Case I If remainder = 0 then check if z is odd (except 5), then last digit = 1 and if z is even then last digit = 6
Case II If remainder = 1, then required last digit = last digit of the base (i.e. z) If remainder = 2, then required last digit = last digit of the base \((z)^2\) If remainder = 3, then required last digit = last digit of the base \((z)^3\)
Note : If z = 5, then the last digit in the product = 5
Example: Find the last digit in (295073)^130
Solution: Dividing 130 by 4, the remainder = 2 Refering to Case II, the required last digit is the last digit of \((z)^2\), ie \((3)^2\) = 9 , (because z = 3)
Great explanation .. kudos _________________
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3^3^3 to solve this we have to take top down approach...we cannot deduce 27^3......it should be 3^27....hence answer is 7
Hey buddy, how do you come from 27^3 to 3^27???? Something here I cant understand...
hi defoue....as i said plz follow top down apprach ...i mean if you have 2^3^4^2 this means 2^(3^(4^2)) hence 2^3^16......2^43046721.......now use our cyclicity table and hence answer is 2...sorry fortaking such weird number but couldn't think of anything else _________________
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Using the cyclisity methodology... : 3^(3^3) = 3^27
Cyclisity of 3 = 4....
27 mod 4 = 3....
Therefore last digit would be 3^3 = 7......
Ans is D _________________
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Re: last digit of a power
31 Jan 2010, 10:51