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Expert Member
Power
Find the last digit of
2^3^4^5^6^7
and
2^3^5^4^5^6^7
logically and not by calculating!!
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Contributing Member
Re: Power
Last digit of both the equation should be 6.
Solution:
In that first equation take the first 3.
2^3 = 8
8^4 = 6
Its very easy to find the last digit of 8*8*8*8.
first multiply 8*8=16. (last digit)6*8=32 (last digit)2*8=16.
Here is the logic..whenever the last digit will come 6, it is sure at the end of any power it should be 6.
In that second equation use the same logic. 2^3^5^4 will contains the last digit 6. so finally we got 6.
i am using this logic..but i think some easiest logic might be there.
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suresh
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Contributing Member
Re: Power
Hey i found one more logic here....
All the 2 power numbers goes to the same order like 2,4,8,6.
So multiply the numbers except 2.
3*4*5*6*7 = 2520.
Divide this number by 4. because we have a same order.
2520/4 = remainder 0. so it should be 6.
Suppose if remainder 1 will come then the answer should 2 and so on.....
Use the same logic u can get the same result for equation 2.
If anyone have other simple logic then post here....
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suresh
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Expert Member
Re: Power
2^3^4^5^6^7 : last digit is 2
2^3^5^4^5^6^7 :last digit is 8
Find the correct logic..other wise i will give it later !!!
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Contributing Member
Re: Power
hi smart coder,
Are you working for that answer...It should be wrong...check it out....
Otherwise you explain..i will find the error....
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suresh
Last edited by psuresh1982; 01-27-2007 at 03:26 AM.
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Expert Member
Re: Power
ok
2^3^4^5^.... = 2^3^(an even number) = 2^(4k +1) ...so the last digit shud be 2.
2^3^5^4.... = 2^3^(an odd number) = 2^(4k+3)... so the last digit shud be 8.
Do U hve ne questions suresh??
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Contributing Member
Re: Power
hi smat coder,
what you mean by k in (4k+1) ?
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suresh
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Contributing Member
Re: Power
yes..yes...i got the logical error in your answer....
ok ....first you explain your solution (question mentioned in my previous mail)...
Then i told you where you made a mistake....
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suresh
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Expert Member
Re: Power
4K+1 means any number of this form like 5,9,13....
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Contributing Member
Re: Power
hi smart coder,
According to the first equation 2^3^4^5^6^7 what is the value of k?
how can you find the value k ?
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suresh
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Expert Member
Re: Power
i cannot find the value of k but i know that 3^(even number) is of type
4k+1 and 2^(4k+1) must end with 2..u dnt require to know the exact value of k!!
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Contributing Member
Re: Power
ok smart coder....i know how the value k coming....
Finally i ask one more question....
According to your logic what is the last digit value for the following equation ?
2^3^4
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suresh
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Contributing Member
Re: Power
so your answer is 2.
ok now we solve it manually.
2^3^4
if we elobarate this one we get 2^3*2^3*2^3*2^3
8*8*8*8 = 4096. so last digit is 6.
Now tell me is it your logic correct ?
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suresh
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Contributing Member
Re: Power
hi smart coder,
see the attached image...your question like that only...
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suresh
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Contributing Member
Re: Power
Here i want all of our puzzle solving friends...
Because i never hear the logic "first you find top one and then finally bottom".
Waiting for our friends reply....Tell me friends which one is correct ?
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suresh
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Expert Member
Re: Power
its 2
2^3^4 = 2^81 = 2^(4k+1)
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Expert Member
Re: Power
ur basics are not correct..while calculating the powers u never go bottom to top..but u come top to bottom!!
2^3^4 is equal to 2^81 not 8^4 !!!
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Expert Member
Re: Power
wt do u think about 2^1^2???.... tell me the last digit...4 or 2
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Expert Member
Re: Power
Kindly answer me Suresh!!!
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Expert Member
Re: Power
i meant this!!!
----4
--3
2
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