#2**+13 **

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

Melody
Mar 27, 2015

#1**+13 **

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

CPhill
Mar 27, 2015

#2**+13 **

Best Answer

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

Melody
Mar 27, 2015