Question of the week 12-Feb-2018

The last two digits of  18^17^16^......^3^2^1 are

a) 68              b) 76             c) 32                d) 24

The question can also be framed as ""What is the remainder when divided with 100?". Answer to the previous week question here. Next week question here. This question is part of my upcoming book and it is solvable in less than 60 seconds. To make it clear, Ex: 4^3^2^1 = 4^9. Your answer only with a logical explanation is welcomed in the comments section. It is still ok for me if you want to google to find the question on internet. You may not find the question as it is self-prepared. Stay tuned for the technique and the answer. I am making sure that my upcoming book is with many such beautiful problems. Reading it will provoke new thoughts and bring out the best in you.

Solution (21-Feb-18):

The last two digits of Powers of 18 follow the sequence of 18, 24, 32, 76, 68, 24, 32, 76, 68, ... and the cycle of 68, 24, 32, 76 continues thereafter. So, last two digits of 18^p are

• 68 if p = 4n+1 (other than very first 18)
• 24 if p = 4n+2 (i.e. 18^2, 18^6, 18^10 ...)
• 32 if p = 4n+3
• 76 if p = 4n.

Our job is done if we can identify the nature of the power (p) of 18. Here p = 17^16^15.....^3^2^1 i.e p = 17^x (Let us call the entire power 'x'). We need to identify what is the remainder of 17^x when divided with 4 (as the above cycle repeats after every four). But, 17^x = (1+16)^x which will clearly leave the remainder 1 when divided with 4 as 16 is divisible by 4. So, the power 17^x is of category 4n+1. So, our required answer is "68".