Power

[smart_coder]

Find the last digit of
2^3^4^5^6^7
and
2^3^5^4^5^6^7

logically and not by calculating!!

[psuresh1982]

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.

----------------
suresh

[psuresh1982]

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....

-------------------
suresh

[smart_coder]

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 !!!

[psuresh1982]

hi smart coder,

Are you working for that answer...It should be wrong...check it out....

Otherwise you explain..i will find the error....

--------------
suresh

[smart_coder]

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??

[psuresh1982]

hi smat coder,

what you mean by k in (4k+1) ?

-----------
suresh

[psuresh1982]

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....

--------------------
suresh

[smart_coder]

4K+1 means any number of this form like 5,9,13....

[psuresh1982]

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 ?

--------------------
suresh

[smart_coder]

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!!

[psuresh1982]

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

------------------
suresh

[smart_coder]

its 2
2^3^4 = 2^81 = 2^(4k+1)

[psuresh1982]

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 ?

---------------
suresh

[smart_coder]

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 !!!

[smart_coder]

wt do u think about 2^1^2???.... tell me the last digit...4 or 2

[smart_coder]

Kindly answer me Suresh!!!

[psuresh1982]

hi smart coder,

see the attached image...your question like that only...


----------------
suresh

[smart_coder]

i meant this!!!

----4
--3
2

[psuresh1982]

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 ?

------------------
suresh