Representing an integer when proving a function?

Question

How do you prove a function when one of the components is only specified as an integer?

For example, how would you prove this function?

Prove: n^5 - n is divisible by 10 for all n is greater than or equal to 2.

Thanks in advance.

Answer 1

Hi,

n^5 - n = n(n^4 - 1)
=n(n^2 + 1)(n^2 - 1)
= n(n + 1)(n - 1) (n^2 + 1)

Let us denote a function UD(n), meaning unit digit or last digit of n, where n => 2

For This to be divisible by 10 the last digit of the reduced expression should be 0, upon substitution of 'n'

Any number n greater than 2 can only end with one of the following unit digits

UD(n) = 0, 1, 2, .....9

case1 : UD(n) = 0

as (n) is a factor of n^5 - n, UD(n^5 -n) = 0
hence n^5 - n is divisible by n, where UD(n) = 0 and where n =>10

Case2: UD(n) = 2, 4, 8
UD[n(n^2 + 1)] = UD(n)*UD(n^2 +1)
= (2^k)*5 = 10.*2^(k-1)..... |2, 4, and 8 are powers of 2 = 2^k
Hence for all numbers n, such that UD(n) = 2, 4 and 8, n^5 - n is divisible by 10

case 3: UD(n) = 1
UD(n - 1) = 0, as (n - 1) is a factor of n^5 - n, for all 'n' ending with 1, n^5 - n is divisible by 10, where n => 11

Case4: UD(n) = 3, 7
UD(n^2 + 1) = 0
As (n^2 + 1) is a factor of n^5 - n, for all n ending with 3 or 7, n^5 - n is divisible by 10

Case 5: UD(n) = 5
UD[n(n + 1)] = 0
As n(n + 1) is a factor of n^5 - n, for all n ending with 5, n^5 - n is divisible by 10

case 6: UD(n) = 6
UD[n(n - 1)] = 0
As n(n - 1) is a factor of n^5 - n, for all n ending with 6, n^5 - n is divisible by 10

Case 7: UD(n) = 9
UD(n + 1) = 0
As (n + 1) is a factor of n^5 - n, for all n ending with 9, n^5 - n is divisible by 10

So the problem statement is proved.

Good luck

Answer 2

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

If n is even, then n is divisible by 2 and (n^4 - 1) is a number divisible by 5. A number that is a multiple of 2 TIMES a number that is a multiple of 5 will be result in a number that is divisible not only by 2 and 5, but divisible by 10 as well. That is how come the number n(n^4 - 1) is divisible by 10....

Now.... if n is "ODD"... then n is NOT divisible by 2.... HOWEVER, the (n^5 - 1) component of n(n^5 - 1) IS divisible by 10.... and that is how come when n is an odd number... the entire n^5 - n number is divisible by 10.... even though the n is ODD... the (n^5 -1) is a number divisible by 10...

Try it out.... you'll see what I'm talking about... =)

Answer 3

Starting with n=2, n^5 always ends with a final digit of n.
Thus when you subtract n from n^5 the resulting number ends in 0 and so is divisible by 10.

Answer 4

n^5 = n mod 5 (fermat's little theorem). Also n^5=n mod 2
since n^2=n mod 2 (also by flt). Conclusion: n^5=n mod 10
as required.

Answer 5

you could coach it via pondering following situations : n could be the two of the type 6k, 6k+a million, 6k+2, 6k+3, 6k+4, 6k+5 if n is of sort 6k then L.H.S. = [2k] +[ok + 4/6] + [ok + 2/6] = 2k+ok+ok = 4k R.H.S. = [3k] +[ok+3/6] = 3k+ok = 4k if n is of sort 6k+a million then L.H.S. = [2k + a million/3] + [ 3k + 5/6] + [ok + 3/6] = 2k+ok+ok = 4k R.H.S. = [3k + a million/2] + [ok + 4/6] = 3k + ok = 4k if n is of sort 6k+2 then L.H.S. = [2k+2/3] + [ 3k + a million] + [ ok + 4/6] = 2k + ok+a million+ok = 4k+a million R.H.S. = [3k + a million] + [ok + 5/6] = 3k+a million+ok = 4k+a million further you could verify for types 6k+3, 6k+4 and 6k+5 for this reason proved