Mathematical Induction: Prove 5^(3n+7) + 6^(2n-1) + 4 is divisible by 5.?

Question

I've actually worked it out quite a bit, but I'm stuck at the end.

Base case:
n = 1: 5^10 + 6^1 + 4 = 5^10 + 2*5 = 5 (5^9 + 2) = 5 | 5 (5^9 + 2)

Assume it's true for:
n = k: 5 | 5^(3k+7) + 6^(2k-1) + 4
Let 5^(3k+7) + 6^(2k-1) + 4 = 5s, s is an integer.

Required to prove:
n = k+1: That 5^(3k+3+7) + 6^(2k+2-1) + 4 is div. by 5.

5^(3k+7)*5³ + 6^(2k-1)*6² + 4
= 5³*5^(3k+7) + (5³ - 89)*6^(2k-1) + 4
= 5³*[ 5^(3k+7) + 6^(2k-1)] - 89*6^(2k-1) + 4
= 5³*[ 5^(3k+7) + 6^(2k-1) + 4] - 5³*4 + 4 - 89*6^(2k-1)
= 5³*5s - 5³*4 + 4 - (90-1)*6^(2k-1)
= 5³*5s - 5³*4 - 5*18*6^(2k-1) + 6^(2k-1) + 4
= 5³*(5s - 4) - 5[18*6^(2k-1)]+ 6^(2k-1) + 4
= 5³*(5s - 4) - 5[18*6^(2k-1)]+ (1+5)^(2k-1) + 4...

And I'm pretty much stuck. I've tried expanding it beyond that, but I just can't seem to get rid of that darned 4 - or at least worm it into another factored 5s, or... anything. Any ideas? Thanks so much in advance!

Answer 1

Base case:
n = 1: 5^10 + 6^1 + 4 = 5^10 + 2*5

so 5 | (5^10 + 2*5)

Assume it's true for:
n = k: 5 | 5^(3k+7) + 6^(2k-1) + 4

Therefore, for n = k+1 we have
5^(3(k+1)+7) + 6^(2(k+1)-1) +4
= 5^(3k+10) + 6^(2k+1) + 4

(n = k+1) - (n = k)
(5^(3k+10) + 6^(2k+1) + 4) - (5^(3k+7) + 6^(2k-1) + 4)
= (5^(3k+10) - 5^(3k+7)) + (6^(2k+1) - 6^(2k-1))

The first term is obviously divisible by 5

take out 6^(2k) on the second term to give

6^(2k)(6 - 6^(-1)) = 6^(2k)(35/6) = 6^(2k-1)*35

which is clearly divisible by 5.

So if the base divisible by 5 and we add only multiples of 5 then in general, 5^(3n+7) + 6^(2n-1) + 4 is divisible by 5.

Answer 2

Prove 5^(3n+7) + 6^(2n-1) + 4 is divisible by 5.?

5^(3n+7) : is divisible by 5 the only factor is 5.

for the other part of the expresssion :
6^(2n-1) + 4

6^any_integer will have a remainder of 1 when divided by 5.

4 + this remainder is divisible by 5

so effective the entire expression is divisbile by 5.
.
if you want to do it your way i think you should only look at the last 2 terms .

for instance :

for n =1 the expression 6^(2n-1) + 4 is divisble by 5

suppose ok for n=1 until n= N-1

now proof that then also for n=N

6^(2N-1) + 4 = 6^(2(N-1) -1 + 2) + 4 =
36*6^(2(N-1) -1 ) + 4 =
35*6^(2(N-1) -1 ) + 6^(2(N-1) -1 )+ 4 =
35*6^(2(N-1) -1 ) + {6^(2(N-1) -1 )+ 4}

first term : 35*6^(2(N-1) -1 ) is divisble by 5 ( because 35 is )
second is {6^(2(N-1) -1 )+ 4} is divisble by 5 because of induction hypothesis.
.bingo.

Answer 3

you have alreadyt proved for base so I need not prove

let it be true of k

5^(3k+7) + 6^(2k-1) + 4 is divisible by k

we need to show that it it is true for k + 1

5^(3(k+1)+7) + 6^(2k+1) + 4 is divisible

we know 1st term is divisible by 5
so we should find diffterence

6^(2k+1) - 6^(2k-1)
= 6^(2k-1)(6^2-1)
= 35*6^(2k-1)

the diffence is divisble by 5 so (k+1)st term

hence proved

Answer 4

do you have chose to recognize each thing step by using step? if so you will greater useful point out that. properly, first you may desire to work out even if if that's real for n=a million, taking L.H.S. n=a million; L.H.S.= 5+3 = 8= 4*2 divisible by using 4. then anticipate that the top result's real for some n=p and you get that 5^p + 3 is divisible by using 4. for this reason could desire to be written as 5^p + 3= 4k (ok€ Z) while n=p+a million; L.H.S.=5^(p+a million) + 3 = 5*5^p +3 =5( 5^p + 3) - 4*3 as this exhibits, 5(5^p + 3) is divisible by using 4 because of the fact we've proved above that 5^p +3 is a assorted of four. and four*3 is divisible by using 4. for this reason while fact is real for n=p, it is likewise maximum appropriate for n=p+a million. by using the theorum of mathematical induction, 5^n + 3 is divisible by using 4 for all n integers.