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How come 30^99 + 61^100 is divisible by 31

Guest Aug 3, 2015

Best Answer 

 #5
avatar+89953 
+10

Here's another way to prove this using the Binomial Theorem

 

Let (30)^99  = (31 - 1)^99

 

Let  (61)^100  = (62 - 1)^100

 

So we have

 

[(99C0)(31)^99  -  (99C1)(31)^98 + (99C2)(31)^97 + ....+ (99C98)(31) - 1 ]

+

[(100C0)(62)^100 - (100C1)(62)^99 + (100C2)(62)^98 + ..... +(100C98)(62)^2 - (100C99)(62) + 1 ]

 

And adding these, the last terms cancel, and every other term in both expressions is divisible by 31.....

 

 

  

CPhill  Aug 3, 2015
 #1
avatar+14536 
+10

(30^99+61^100)  is divisible by 31 !

(30^99+61^100) modulo 31 = 0

radix  Aug 3, 2015
 #2
avatar+14536 
+5

radix  Aug 3, 2015
 #3
avatar+93656 
+5

Good thinking Radix.   

I would like to see a proof though.  :/

Melody  Aug 3, 2015
 #4
avatar+20025 
+10

How come 30^99 + 61^100 is divisible by 31

 

$$\small{\text{$30 \equiv - 1 \pmod {31}$}}\\
\small{\text{ and $61 \equiv - 1 \pmod {31}$}}\\\\
\small{\text{$(-1)^{99} + (-1)^{100} \stackrel{?}\equiv 0 \pmod{31}$}}\\\\
\small{\text{$-1 + 1 \equiv 0 \pmod{31}$}}\\\\$$

 

heureka  Aug 3, 2015
 #5
avatar+89953 
+10
Best Answer

Here's another way to prove this using the Binomial Theorem

 

Let (30)^99  = (31 - 1)^99

 

Let  (61)^100  = (62 - 1)^100

 

So we have

 

[(99C0)(31)^99  -  (99C1)(31)^98 + (99C2)(31)^97 + ....+ (99C98)(31) - 1 ]

+

[(100C0)(62)^100 - (100C1)(62)^99 + (100C2)(62)^98 + ..... +(100C98)(62)^2 - (100C99)(62) + 1 ]

 

And adding these, the last terms cancel, and every other term in both expressions is divisible by 31.....

 

 

  

CPhill  Aug 3, 2015
 #6
avatar+93656 
0

Great answers.  Thanks Chris and Heureke.

I especially like yours Heureka. 

I also like that Latex stacked question mark.

That'll have to find its way to the latex thread :)  I have added it  now :)

Melody  Aug 3, 2015

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