I am still getting used to these questions and I still like them. :)
I'd do it the same as CPhill but I would start with a smaller number.
$$\\7^1=7\\
7^2=49\\
7^3=343\\
7^4=2401\\
$that is a really helpful one$\\
$(Any number ending is 01) raised to any positive integer n must also end in 01 because only the last 2 digits will affect the outcome and 1^n=1 $ \\\\
(7^4)^{n} $ must end in 01 for any positive integer n$\\\\
7^{2014}=7^{2012}*7^2\\
7^{2014}=7^{4*503}*7^2\\
7^{2014}=(7^4)^{503}*49\\
7^{2014}= ....01 *49\\
7^{2014}= ....49\\$$
so the last 2 digits will be 49 Just like CPhill said
Notice that
7^20 ends in 01
So
(7^20)^5 = 7^100 also ends in 01
And
7^2000 = (7^100)^20 will also end in 01
And 7^14 ends in 49
So 7^2014 = 7^2000 x 7^14 = .........01 x ........49
Will also end in 49
I am still getting used to these questions and I still like them. :)
I'd do it the same as CPhill but I would start with a smaller number.
$$\\7^1=7\\
7^2=49\\
7^3=343\\
7^4=2401\\
$that is a really helpful one$\\
$(Any number ending is 01) raised to any positive integer n must also end in 01 because only the last 2 digits will affect the outcome and 1^n=1 $ \\\\
(7^4)^{n} $ must end in 01 for any positive integer n$\\\\
7^{2014}=7^{2012}*7^2\\
7^{2014}=7^{4*503}*7^2\\
7^{2014}=(7^4)^{503}*49\\
7^{2014}= ....01 *49\\
7^{2014}= ....49\\$$
so the last 2 digits will be 49 Just like CPhill said