#5**+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

#4**+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**+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