Thread: Power

lemme tell u wt i askd was based on a CAT question!!
nd do u want to challenge IIMs??

Well smart coder....
i want to clarify myself...Because "Nobody knows all the things".
That's why i need a help for other people.

Also it is very difficult to find the value for your logic...
I need a clear explantion about the 2^(4k+1).

In my next example you simply told 2^81. so you calculate the 81 like 3*3*3*3. it is single power so you can calcualte easily...if it is large then how can you find ?

If you explain this one i accept that. Waiting for your reply...

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suresh

I am not getting ur point...try to put another example!!
you want other people's view...I also agree to that..beacuse we have many intellectuals present in this forum..and I want to invite everyone to read the whole thread and then comment.
But what I asked is what I learnt at Career Launcher(a reknowned coaching for MBA) and if u ask this question to a person who is preparing for CAT will give the same answer as mine.
Thanks

wt wt wt anushya?
I m totally confused by ur question...plz write clearly!!

ok smart coder...

Explain the following one...

This is you already posted....

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.

In that above you mentioned 2^3^(an even number)..

how you told that is an even number and how you write 2^(4k+1).....Also how the value 1 came that place ?

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

Yes anushya ...the logic is defined already that u come top to bottom..
and suresh this is for you:

4^(any number) is an even number
so 3^4^(any number) = 3^(even number)

now 3^(even number) = 4k+1 type

like 3^2 = 9 = 4*2 + 1
and 3^4 = 81 = 4*20 + 1

ne other doubts?
please feel free to ask..

anushya this is for you..tell me the last digit for

2^1^2

if it is 3^(even) then 4K+1
or if 3^(odd) then 4K+3

hi smart coder,

Thanks for your all the information....This is the first time i hear these type of concept....I want to know the concept..that's why i ask so much of questions...now i have some idea to solve these type puzzle...still i have one more doubt...

you told me if it is 3^(even) then 4K+1.
suppose some one give the equation...2^7^6^9^4 or simply 7^6^9^4.

how can i solve this equation ? explain this also...

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

look there can't be answers for all such type questions...u can have answers if u think very logically
for ur question:7^6^9^4 i have a solution:
6^(any number) is of form 4k except 6^1
in ur question it is obvious dat it 6^(a large number) so it will be of type 4k
Now 7^(4k) will always end with a 1.

For u anushya:
u can c many problems when brackets are not applied but d meaning is applied like the following examples:

--4
-3
2
this is not equal to

-12
2


u cannot multiply the powers until the brackets are told explicitly..

Yeah Smart coder i agree with u...

2^3^4 is equal to 2^81....

if there is no bracket then the concept is top to bottom approach... Here first we need to find 3^4 and then 2^81

if we want to find bottom to top approach then we should use bracket its mandatory...

(2^3)^5 = 2^(3+5)

Thanks
Manoj

i knew dat atleast manoj will support me..i was waiting for u only!!!

i think my this problem wud hve cleared basic mathematics concepts of many viewers!!
dats y i put this in!!

What more you want to know anushya??
what more working process??
the process is simple
just TOP to BOTTOM!!!
dats it!!
feel free to ask more!!