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Find the last 2 digits of 7^2014?

Guest Mar 27, 2015

Best Answer 

 #2
avatar+92775 
+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
avatar+87294 
+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
avatar+92775 
+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

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