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I'm trying to help a friend with math...
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You are currently reading a thread in /sci/ - Science & Math

I'm trying to help a friend with math study, she has an induction proof question that she needs help with. I haven't done anything like this in 7 years so I'm rusty, can someone please help with a proof?
>>
I feel like I'm overcomplicating this...
>>
k=0 works, assume k=n works, plug in k=n+1 and factor 11 out
>>
>>7855452
Case 1: K is even so 5k=10n
5^10n+1 + 4^10n+2 + 3^10n = 5+16+1=22=0 mod 11
Case 2: K is odd so 5k=10n+5
5^6 + 4^7 + 3^5 = 3^3 + 5*4^2 + 4*3 = 6 + 4 + 1 = 0 mod 11

QED
>>
>>7855537
Terrible. Use induction, you idiot.
>>
>>7855537

this is not correct, use induction
>>
OP here

I know the method of induction, but I get stuck when subbing in k=n+1, I don't see how to go from 5^(5n+2)+4^(5n+3)+3^(5n+1) to anything productive.

Am I on the right track if I separate it into, say 5^(5n+1)×5+4^(5n+2)×4+3^5n×3?
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>>7856722
Put n+1 in for k you nerd burglar.
>>
>>7856722

here's a hint

for induction you suppose that

5^(5*k + 1) +
4^(5*k + 2) +
3^(5*k)
% 11
=
0

it also holds that

(5^(5*k + 1) % 11 +
4^(5*k + 2) % 11+
3^(5*k) % 11)
% 11
=
0
>>
16=5 mod 11, 5^5=25·25·5=3·3·5=45=1 (mod 11), 4^5=16·16·4=5·5·4=3·4=1 (mod 11), and also 3^5=81·3=4·3=1 (mod 11).

Hence, 5^{5k+1}+4^{5k+2}+3^{5k}=5·5^{5k}+16·4^{5k}+3^{5k}=5·(1^k+1^k)+1^k=5·2+1=11=0 (mod 11).

Therefore for any natural k, the original sum is divisible by 11.
>>
>>7856722

also you did the n+1 case wrong, should be

5^(5*(n+1) + 1) +
4^(5*(n+1) + 2) +
3^(5*(n+1))

=

5^(5*n + 1 + 5) +
4^(5*n + 2 + 5) +
3^(5*k + 5)

=

5*5 * 5^(5*n + 1) +
4^5 * 4^(5*n + 2) +
3^5 * 3^(5*n)
>>
x[k] = 5^(5k+1) + 4^(5k+2) + 3^5k

Replace k with k+1, resulting in

x[k+1] = 5^(5k+1+5) + 4^(5k+2+5) + 3^(5k+5)
= 5^5 * 5^(5k+1) + 4^5*4^(5k+2) + 3^5*3^5k
= 3125*5^(5k+1) + 1024*4^(5k+2) + 243*3^5k
= 3124*5^(5k+1) + 1023*4^(5k+2) + 242*3^5k + x[k]

3124 = 284*11
1023 = 93*11
242 = 22*11
>>
Oh shit I can't believe I screwed up the expansion. Thanks guys
>>
>>7855452
5^5 = 3125 = 1 mod 11
so 5^5k = 1 mod 11
and 5^(5k+1) = 5 mod 11

4^5 = 1024 = 1 mod 11
so 4^5k = 1 mod 11 and 4^(5k+2) = 5 mod 11

3^5 = 1 mod 11
so 3^5k = 1 mod 11

5^(5k+1)+4^(5k+2)+3^5k = 5+5+1 mod 11 = 0 mod 11
qed

this is too easy for an engineer.
>>
>>7856712
it is

>>7855793
induction is longer and unneeded.
>>
>>7856789
>it is

you're a retarded faggot, go back to /b/
>>
>>7855793
>>7856712
His proof is correct and better than induction since it uses Fermat's first theorem.
Plebs.