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Any ideas?

Prove that 3^4^5+4^5^6 is a product of two integers, each of which is larger than 10^2002.
Tried a couple of things but can't find a reasonable approach.
Reply 1
Original post by ValerieKR
Prove that 3^4^5+4^5^6 is a product of two integers, each of which is larger than 10^2002.
Tried a couple of things but can't find a reasonable approach.


Is this (34)5(3^4)^5 or 3453^{4^5}?
Reply 2
Original post by Zacken
Is this (34)5(3^4)^5 or 3453^{4^5}?


2nd one
Reply 3
Original post by ValerieKR
2nd one


Write 345+456\displaystyle 3^{4^5} + 4^{5^6} as m4+4n4m^4 + 4n^4 where you choose a suitable value for nn, then factorise and show the smaller factor is greater than 10^(2002).
Reply 4
Original post by Zacken
Write 345+456\displaystyle 3^{4^5} + 4^{5^6} as m4+4n4m^4 + 4n^4 where you choose a suitable value for nn, then factorise and show the smaller factor is greater than 10^(2002).


Thanks

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