Number Theory

[gauravkukreja]

WHich is greater?
31^11 or 17^14

:bugeye:

[Galileo]

31^11 = 25408476896404831
17^14 = 168377826559400929

So 17^14 is biggah

[shmoe]

without a calculator or computer, notice that (in white):


31^11<(2^5)^11<2^55<2^56<(2^4)^14<17^14

[Noticibly F.A.T]

hey shmoe i don't get it

[TenaliRaman]

another brilliant work of shmoe there,
(check between :: and ::)
::
31 < 32 .... (Order of Natural Numbers -- Peano's Axiom)
31 < 2^5 ... (since 2^5 = 32)
31^11 < (2^5)^11 ... (Raising both sides to power of 11)
31^11 < 2^55 ... (Laws of indices) .. *

But,
2^55 < 2^56 ... (Laws of indices) .. **

therefore from * and **,
31^11 < 2^56 ... (Transitivity)
31^11 < 2^(14*4) ... (bcos 14*4 = 56)
31^11 < (2^4)^14 ... (Laws of indices)
31^11 < 16^14 ... (2^4 = 16) .. ***

But,
16 < 17 (Order of Natural Numbers -- Peano's Axioms)
16^14 < 17^14 ... (Raising both sides to the same power of 14) .. ****

therefore from *** and **** we get,
31^11 < 17^14 .. (Transitivity)
::

-- AI

[Gokul43201]

That is quite nice.

I did it the clumsy way...albeit using only paper and pencil...and got lucky !

Compare 11*log31 and 14*log17

This takes a lot longer to write down here than to actually do...

log 30 = 1 + log3 = 1.477 and log 33.3 ~ 2 - log3 = 1.523. A rough linear interpolation tells me that log31 ~ 1.49. So, 11*log31 ~ 16.4

log17 ~ log (100/6) = 2 - log3 - log2 = 2 - 0.778 ~ 1.22, so 14*1.22 > 14*1.2 = 16.8, which is the bigger of the two.

PS : It helped that I know the logarithms of 1, 2, 3, 5, and 7...you can get most others pretty quickly from these.

[Noticibly F.A.T]

Now i get it. Thanks tenaliraman