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What are the last two digits of 5^2015

Guest May 19, 2015

Best Answer 

 #2
avatar+14536 
+19

Here are all digits !

Gruß radix !

radix  May 19, 2015
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6+0 Answers

 #1
avatar
+5

25 detail detail detail detail

Guest May 19, 2015
 #2
avatar+14536 
+19
Best Answer

Here are all digits !

Gruß radix !

radix  May 19, 2015
 #3
avatar+631 
+3

∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity

thejamesmachine  May 19, 2015
 #4
avatar+90988 
+5

I aM going to think about this

 

5^2015

 

$$\\5^1=5\\
5^2=25\\
5^3=125\\
5^4=525\\$$

 

It seems that if the power is bigger than 1 that the last 2 digiits will always be 25

MAYBE SOMEONE WOULD LIKE TO PROVE THIS  ??

Melody  May 20, 2015
 #5
avatar+78577 
+10

Here's a "proof"

5^2 = 25

5^3  =  25 * 5  =  (20 + 5) (5) = 100 + 25

5^4  = 125 * 5  = (120 + 5) (5)  = (100 + 20 + 5) (5) = 500 + 100 + 25

5^5 = 625 * 5 =  (600 + 20 + 5) (5)  = (500 + 100 + 25)(5) = (500 + 100 + 20 + 5) (5)  = 2500 + 500 + 100 + 25

 

So it appears that the pattern  for 5^n = 

 

25 + 100*5^(0) + 100*5^(1) +.....+ 100*5^(n - 4) + 100*5^(n- 3)  for n ≥ 5 

 

And, in this series, all terms except the first one will end in the digits "00"...so adding 25 to this sum will always result in a number ending in "25"..........

 

CPhill  May 20, 2015
 #6
avatar+90988 
0

Thanks CPhill  

Melody  May 21, 2015

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