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

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

Aug 3, 2015

#5
+115955
+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.....

Aug 3, 2015

#1
+14537
+10

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

Aug 3, 2015
#2
+14537
+5

#3
+112463
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I would like to see a proof though.  :/

Aug 3, 2015
#4
+25639
+10

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

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

Aug 3, 2015
#5
+115955
+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
#6
+112463
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 :)

Aug 3, 2015