Question

Prove by induction 5^(n+1)+4x6^n when divided by 20 leaves the remainder 9 for all n element of N

Answer 1

\[5^{n+1}+4.6^{n} = 20 m + 9 \]
\[5^{n+1+1}+4.6^{n+1} = 5.5^{n+1}+6.4.6^{n} = 5.5^{n+1}+ (5+1)4.6^{n} = 5.5^{n+1}+ 5.4.6^{n} +4.6^{n} \]
\[ = 5[5^{n+1}+4.6^{n}] + 4.6^{n} = 5[20m+9] +4.6^{n} \]

Answer 2

can we thus say that p(n+1) also leaves a remainder 9 when divided by 20 ?? else pls show me how to present the answer !!

Answer 3

are those dots multiplication?

Answer 4

yes

Answer 5

\[\times\]

Answer 6

.\times

Answer 7

ok let me look at this for a sec

Answer 8

yes the dots indicate multiplication

Answer 9

hmm, I dont see how this would imply that it leaves a 9 as a remainder.

Answer 10

you never actually use the inductive hypothesis, I see that you plugged it in, but notice that it actually does nothing, as what you replaced it with would still be an integer.

Answer 11

i want to know how to prove the statement by induction..i was just trying it ... i know i was not correct .. in case you can help with this ?

Answer 12

im trying to figure it out as well. I took the same steps as you and arrived at the same point.

Answer 13

ok got it

Answer 14

5*4*6^n+4*6^n+5*5^(n+1)

Answer 15

so
20*6^n+4*6^n + 5*5^(n+1)

Answer 16

20*6^n+20m+9 = 20(6^n+m)+9

Answer 17

understand?

Answer 18

how did you replace 4*6^n + 5*5^(n+1) = 20m+9 as
5^(n+1)+4x6^n = 20m+9 ???

Answer 19

you have
\[5*5^{n+1}+6*4*6^n \]this is in your second line
so
\[5*5^{n+1}+6*4*6^n = 5*5^{n+1}+5*4*6^n+4*6^n = 20*6^n+(5*5^{n+1}+4*6^n)\\\text{by IH} = 20*6^n+20m+9 = 20(6^n+m)+9\]

Answer 20

that is what i am trying to point out 5∗5^(n+1)+4∗6^n means 5^(n+2) + 4*6^n which is not equal to 20m+9

Answer 21

oh I wrote something down wrong, sorry

Answer 22

By IH, \(5^{n+1}+4.6^{n} = 20 m + 9\)
not \(\color{red}{5} * 5^{n+1}+4.6^{n} = 20 m + 9\)

?

Answer 23

yes correct !!

Answer 24

well im guessing ganesh has it:) so im back to my movie

Answer 25

ohk.. lets start over ive something but to prove n+1, but it uses congruences... can we use them ?

Answer 26

which movi u watching zz ? :)

Answer 27

about to start Synecdoche, New York (2008)

Answer 28

it is a problem given in a class 11 book, where congruences have not been discussed

Answer 29

oh then let me think, however just to prove that this is actually true :-
\(5[5^{n+1}+4.6^{n}] + 4.6^{n} \equiv 5*9 + 4.6^n \equiv 5 + 4*6^n \equiv 9\mod 20\)

\(5 + 4*6^n \equiv 9\mod 20 \)
\( 4*6^n \equiv 4\mod 20 \)
\( 6^n \equiv 1\mod 5 \)
By FLT its true. QED

So we're pursuing a legitimate proof, but this is not acceptable at 11th grade. let me think a bit...

Answer 30

for 11th grade, id slightly change the above arguemnt by hiding congruences :

\(5[5^{n+1}+4.6^{n}] + 4.6^{n} \)

\(5[20m+9] + 4.6^{n} \)

\(5[20m+9] + 4(5k+1)\) *****

\(5*20m+ 45 + 20k+4\)

\(20(5m +k+ 2)+9\)

Answer 31

how did you get 4.6^n = 4(5k+1) ?

Answer 32

***** we can prove it easily

a sub induction proof for showing :
\(6^n = 5k+1\)

or in words, \(6^n\) leaves a remainder of 1, when divided by 5

Answer 33

you want me prove... or you wana give it a try.

im not sure if there is any we can avoid dealing wid this

Answer 34

ok i got it .. thanks

Answer 35

good :) proving that 6^n-1 is divisible 5 is easy, give it a try...

Answer 36

sure :)