Is n5-n divisible by 30 for all integral values of n >1. Justify.?

Answer 1

I'm sure you mean n^5 - n.

n^5 - n
= n(n^4 - 1)
= n(n² - 1)(n² + 1)
= n(n - 1)(n + 1)(n² + 1)

The first three terms n, n - 1, and n + 1 represent three consecutive integers. So at least one of those integers must be divisible by 3. Also, at least one of them must be even. So n^5 - n must be divisible by 6...now we just have to find a way to show that at least one of the factors is divisible by 5.

If any of n, n - 1, or n + 1 is divisible by 5, then the product must be divisible by 5, and therefore n^5 - n is also divisible by 6 x 5 = 30, from the last paragraph.

If none of n, n - 1, and n + 1 is divisible by 5, then n mod 5 must be 2 or 3. (This means that, when divided by 5, n must have a remainder of 2 or 3. So, for example, if n were 7 (7 mod 5 = 2), n - 1 is 6 and n + 1 is 8, and none of them is divisible by 5. For another example, if n were 13 (13 mod 5 = 3), n - 1 is 12 and n + 1 is 14, and none of these is divisible by 5.)

In the case of n mod 5 = 2, we can write n as 5k + 2. Then n² + 1 = (5k + 2)² + 1 = 25k² + 20k + 4 + 1 = 25k² + 20k + 5 = 5(5k² + 4k + 1), which is clearly a multiple of 5.

In the case of n mod 5 = 3, we can write n as 5k + 3. Then n² + 1 = (5k + 3)² + 1 = 25k² + 30k + 9 + 1= 25k² + 30k + 10 = 5(5k² + 6k + 2), which again is clearly a multiple of 5.

Therefore, no matter what, n^5 - n is divisible by 30.

Answer 2

enable's denote (a million) An = n^5 - n we are in a position to factorize 30 like this (2) 30 = 2 * 3 * 5 So we ought to proove that n^5 - n is divisible via 2, 3 and 5 enable us to first rewrite n^5 - n like this (3) n^5 - n = n(n^4 - a million) = n(n^2 - a million)(n^2 + a million) = n(n - a million)(n + a million)(n^2 + a million) Divisibility via 2 -------------------- An is divisible via 2 despite if this is a good quantity. If n is even then (3) yields that An is even (because of the fact in case you multiply any quantity with a good quantity you will get a good quantity) If n is strange, n - a million is even and (3) yields that An is even This prooves that An is divisible via 2 Divisibility via 3 ------------------- To proove that An is divisible via 3 we ought to proove that the two n, n - a million, n + a million or n^2 + a million is divisible via 3 if n = 3k, then An is divisible via 3 If n = 3k + a million then n - a million = 3k which yields that An is divisible via 3 If n = 3k - a million then n + a million = 3k which yields that An is divisible via 3 Divisibility via 5 ------------------- we will take the simmilar mindset as we did in divisibility via 3 we will attempt to coach it in 4 distinctive situations. If n = 5k, then An is divisible via 5 If n = 5k + a million then n - a million = 5k which yields that An is divisible via 5 If n = 5k - a million then n + a million = 5k which yields that An is divisible via 5 If n = 5k + l, the place l = -2, 2 then neither n nor n - a million nor n + a million are divisible via 5, so we ought to proove that n^2 + a million is divisible via 5. enable's positioned n = 5k + l into n^2 + a million we get that n^2 + a million = (5k + l)^2 + a million = 25k^2 + 10kl + l^2 + a million We see that the contributors via ok^2 and ok are divisible via 5 we ought to proove that l^2 + a million is divisible via 5 for l = -2, 2. yet considering that l^2 = 4 for the two l = -2, 2, we get tath l^2 + a million = 5 for the two l = -2, 2, which yields that l^2 + a million is divisible via 5 for the two l = -2, 2 as a result An is divisible via 5 So An is divisible via 30 for each n > a million while you're taking into the honour that 0 is divisible via any quantity then n^5 - n is divisible via 30 for any n > 0

Answer 3

Lets suppose n5 means n^5 which means n*n*n*n*n.

In this case
n^5-n = n*(n^4-1) = n*(n^2+1)*(n^2-1) = n*(n^2+1)*(n+1)*(n-1).
[note that I used twice (a+b)*(a-b) = a^2-b^2]

This product is divisible by 30 is it is divisible by all the factors of 30 which are: 2, 3, 5.

The product certainly divisible by 2 for any n, because either n or n+1 is divisible by 2.
The product certainly divisible by 3 for any n, because either n-1 or n or n+1 is divisible by 3.
The product certainly divisible by 5 for any n, because any n can be written in one of the following forms:
5k-2
5k-1
5k
5k+1
5k+2
Where k is some integer. For example n=21 can be written as 5*4+1, in this case k = 4, but it is not important.
So.
If n can be written in the form of n=5k-1, then n+1 is divisible by 5 so the product is divisible as well.
If n can be written in the form of n=5k+1, then n-1 is divisible by 5 so the product is divisible as well.
If n can be written in the form of n=5k, then n is divisible by 5 so the product is divisible as well.
If n can be written in the form of n=5k+2 or 5k-2, then n^2+1 = (5k+/-2)^2+1 = 25k^2 +/- 10k + 4 + 1 = 5*(5*k*k+/-2*k+1) which is divisible by 5. [note that I used (a+b)^2 = a^2+2*a*b+b^2]

So. The product is divisible by 2,3 and 5 for any n, so it is divisible by 30 as well.

ps.: for n=1, n^5-n = 1-1 = 0 which is divisible by 30.
for n = 2 n^5-n = 32-2 = 30, OK.
for n=3 n^5-n = 243-3 = 240, OK.

Vote for me!

Answer 4

No. If n=1 the result is 4 which is not divisible by 30.

Answer 5

I reckon no - take n=2 then 25-5=20 which is not divisible by 30 - or if you meant n*5-n then10-5=10 which ain't divisible by 30 either

Answer 6

Yes because 2^5=32-2=30. 30 is divisible by 30.
25^5=9765625-25=9765600. That divided by 30=325520

Answer 7

Yes it can be justified by mathematical induction

for n=k

k^5-k/30

for k=1

1-1/30=0 therefore not valid for n=1

for k=2

(2^5-2)/30=1 which is divisible.Hence true

for k=3

(3^5-3)/30=8

and so on

let f(n+1)=(k+1)

(k+1)^5-(k+1)=k^5+5k^4+10K^3+10k^2+5K+1-k-1

for f(n)=k

k^5-K

so f(n+1)-f(n)
k^5+5k^4+10k^3+10k^2+5k+1-k-1-k^5+k

5k^4+10k^3+10k^2+5k

5(k^4+2k^3+2k^2+k) which is divisible by 30

gives out (k^4+2k^3+2k^2+k)/6

hence its true for all values of n>1



5(k^4+2k^3+2k^2

Answer 8

well yea any number is divisble by 30 it will just give you a decimal value.

Answer 9

no.Try the trial and error method

Answer 10

what the heck is "n5"?