top button
Flag Notify
    Connect to us
      Site Registration

Site Registration

Is it possible to write 3^2016 + 4^2017 as the product of two numbers, both of which are over 2018^183?

+2 votes
159 views
Is it possible to write 3^2016 + 4^2017 as the product of two numbers, both of which are over 2018^183?
posted Feb 20, 2018 by anonymous

Share this puzzle
Facebook Share Button Twitter Share Button LinkedIn Share Button

1 Answer

0 votes

Essentially it is to be proven that
3^2016 + 4^2017 > (2018^183)^2
3^2016 + 4^2017 > 2018^366
4^2017 = 2^4034
2048 = 2^11
2048^366 = 2^(11*366) = 2^4026
Now
(4^2017 = 2^4034) > (2048^366 = 2^4026) > 2018^366
Therefore
4^2017 > 2018^366
So its only logical that
4^2017 + 3^2016 > 2018^366.

answer Feb 22, 2018 by Tejas Naik



Similar Puzzles
+1 vote

Is it possible to write down 1,2,3..100 in some order (one after an other), such that the sum of any two adjacent numbers is a prime number?

0 votes

You have 2017 cards numbered 1, 2, 3, ..., 2017 in the same order. In each move, you can change the order of two adjacent cards. (For example, if you had only four cards arranged as 1234, what you could achieve in one move would be 2134, 1324, or 1243.)

What is the minimum number of moves required for the above cards to be arranged backwards as 2017, 2016, ..., 2, 1?

...