## The Last Two - 19/06/2013

### The Last Two - 19/06/2013

Can you calculate/predict the last two digits of number A?

A=(207^{19}-41)^{10}

showmyiq

Posts: 390
Joined: Sat Sep 15, 2012 9:45 pm

### Re: The Last Two - 19/06/2013

24

(207^19-41) find last two digits of that first.

(7^19-41)

7^19 last two digits = 43

(43-41) = 2

2^10 = 1024

24
jahjaylee

Posts: 1
Joined: Wed Jun 19, 2013 7:35 pm

### Re: The Last Two - 19/06/2013

You are the one who solved the puzzle first!
I will add some further explanation for the other users, who are interested in the approach of solving such problems in future.

(207^{19}-41)^{10}\equiv r \pmod {100} \\
\text{We have that: } 207\equiv 7 \pmod {100} \\
\text{Therefore: } 207^3\equiv 7^3 = 343 \equiv 43 \pmod {100} \\
207^4 = 207^3*207 \equiv 43*7 = 301 \equiv 1 \pmod {100} \\
207^{16} \equiv 1 \pmod {100} \\
207^{19} \equiv 43 \pmod {100} \\
207^{19} - 41 \equiv 2 \pmod {100} \\
(207^{19} - 41)^{10} \equiv 2^{10} = 1024 \equiv 24 \pmod {100} \\

showmyiq