maths questions?

Question

1) Prove that no integer in the sequence 11, 111, 1111, 11111... is a perfect square.
2) If n is an odd integer, show that n^4 +4n^2 +11 is of the form 16k.
3) By using the division alogirthm, show
(i) if a and b are odd integers, then 8|(a^2-b^2)
(ii)if p>3 is a prime, then 24|(p^2-1)
(iii) for n>1, prove that n(n+1)(2n+1)/6 is an integer.
4) Find the largest integer n such that n is a divisor of a^5-a for all integers a.

Answer 1

1) In order to end in a 1 , the number being squared would have to end in a 1 or a 9. The tens digit for the square of any number ending in 1 follows the pattern 0,2,4,6,8,0,2,4,6,8,... so it never gives a number with all ones as the digits.
1^2=01 11^2=121 21^2=441 31^2=961 41^2=1681 51^2=2601

The 10s digit for the square of any number ending in 9 follow the pattern that they are 8,6,4,2,0,8,6,4,2,0, etc, so again no number with all ones will ever be a perfect square.as well.
9^1 = 09 9^2 = 81 19^2 = 361 29^2=841 39^2=1521 49^2=2401

So no number in that sequence will ever be a perfect square.

2. Lets pick any number, x. If we multiply that by 2 then we know 2x is an even number because it's divisible by 2. Therefore a number that is 1 more than the even number 2x is the odd number 2x + 1. So, for our proof we will say that we will change n to 2x + 1 as something that will always be an odd number for any integer value of x. If we substitute 2x + 1 in place of n into
n^4 + 4n^2 + 11, it becomes (2x + 1)^4 + 4(2x + 1)^2 + 11. This multiplies out to 16x^4 + 32x^3 + 24x^2 + 8x + 1 + 16x^2 +16x + 4 + 11 = 16x^4 + 32x^3 + 40x^2 + 24x + 16. Since you are trying to prove that it is of the form 16k, you want to prove 16 divides into everything evenly. Three terms are easy, since 16 and 32 are divisible by 16 alredy. The harder terms are 40x^2 + 24x. We can split that apart into 16x^2 + 24x^2 + 24x. Then the 16x^2 is divisible by 16, so all we have left to prove is that 24x^2 + 24 x is divisible by 16. Before we go on, notice that 16 is really 2^4, So in order to be divisible by 16, 24x^2 + 24x must be divisible by four 2s. To show this, factor 24x^2 + 24x into 24x( x +1). 24 factors on down into 3 * 2 * 2 * 2. There are 3 of the 4 twos that we need to divide by 16, but we still need one more 2 somewhere. Fortunately we are left with x(x + 1) and since those represent 2 consecutive numbers, one of them will be odd and the other will be even. That means the even one has to be divisible by the fourth 2. Therefore 16x^4 + 32x^3 + 40x^2 + 24x + 16 is of the form 16k.

If you want to double check this on a graphing calculator, put 16x^4 + 32x^3 + 40x^2 + 24x + 16 into Y1 and (16x^4 + 32x^3 + 40x^2 + 24x + 16)/16 into Y2. Go to table set and make your first number 1 and your deltaTable = 2 so your input values are all odd. Then, if you look at the values in the table, you will see that they are all integer values for Y2, showing that 16 divides evenly
into every value you get.

3i If a and b are odd, then a^2 and b^2 are also odd. When you subtract and odd number from another odd number, the result is an even number. Therefore a^2 - b^2 is even. Therefore, 8 is divisible by a^2 - b^2, or the fraction can at least be reduced because both numbers are even.

3ii If P> 3 is prime, then it must be an odd number. An odd number squared is still odd, BUT that odd number minus 1 gives an even number. Therefore, 24 is divisible by (p^2 - 1), or the fraction can at least be reduced because both numbers are even.

3iii In order for n(n+1)(n+2)/6 to be an integer, n(n+1)(n+2) must be divisible by 6. Since 6 factors into 2 times 3, in order for a number to be divisible by 6, it must be divisible by both 2 and 3.
If n>1, then values for n, n + 1, and n + 2 must be 3 consecutive integers. Some values coud be:
n n + 1 n + 2
2 3 4 < divisible by 2 and 3
= =


3 4 5 < divisible by 2 and 3
= =

4 5 6 < divisible by 2 and 3
= =

This shows that out of 3 consecutive numbers, one of them has to be divisible by 2 and one has to be divisible by 3. Therefore,
n(n+1)(n+2) would be divisible by 6.

4) For all integers a, a^5 - a factors into:
a^5 = a( a^4 -1) = a(a^2 - 1)(a^2 + 1)

First a^2 is always positive (or zero). But "a" could be either positive or negative. In order to find the largest integer value of a divisor n, it must be positive. So, of these 3 factors, a^2 + 1 and a^2 -1 are the only ones that are nonnegative. That means that n could be equal to the product of these 2 positive factors.
So, n = (a^2 - 1)(a^2 + 1) or a^4 - 1.

Answer 2

1) each item the sequence 11, 111, 1111, 11111... Can be written as the series 1+10^1+10^2+10^3++………+10^N-1

i)If N is Odd: the item =10^(N-1)+(1+1+10^1+10^2+10^3+…….10^(N-2)
10^(N-1) is perfect square of the Number 10^((N-1)/2)
the nearest greater number after this root is 10^((N-1)/2)+1
But this number squared gives((10^((N-1)/2)+1)^2=10^(N-1)+1+2*(10^((N-1)/2)
Which is greater than the term =1+10^1+10^2+10^3++………+10^N-1
Then the series 1+10^1+10^2+10^3++………+10^N-1 is not a perfect square

ii)If N is Even: the item 1+10^1+10^2+10^3++………+10^N-1 can be written
as (10^N-1)/(10-1) =(10^N-1)/9
But 10^N is a perfect square of the Number 10^(N/2) and 9 is a perfect square to 3
Then 10^N-1 is not a perfect square and (10^N-1)/9 is neither

From i,ii
There is no perfect squares in sequence 11, 111, 1111, 11111...


2)Let (n)=2m+1 where m = 0 ,1,2,3,4,,,,,,,,,
n^4 +4n^2 +11=16m^4+32m^3+24m^2+8m+1 +16m^2+16m+4+11
= 16m^4+32m^3+40m^2+24m+16
=16m^4+32m^3+8m(5m+3)+16
If m is Odd 5m+3 is even
If m is even 8m mode 16=0
8m(5m+3) mode 16=0 at all values of m
16m^4+32m^3+8m(5m+3)+16 mode 16=0
hence
(n^4 +4n^2 +11)mode 16=0 at n is any Odd number
n^4 +4n^2 +11 is of the form 16k where k is a whole number if n is Odd number

Answer 3

1.if any of them is a perfect square it should be the square of a number ending in 1 or 9 in which case the number in the 10 s place cannot be 1 and so none of them can be a perfect square
2.n^4+4n^2+11
true for n=1
let it be true for some 'r'
if it is true for r+2then it holds by induction
[(n+2)^4-n^4]+4[(n+2)^2-n^2]=[n^2+4n+4+n^2][n^2+4n+4-n^2]
+4(n+2+n)(n+2-n)=[2n^2+4n+4][4n+4}+4(2n+2)(2)
you find that all the terms are multiples of 16 and hence proved
3.i am afraid the sum is incomplete

Answer 4

I wonder why you asked such a question is. Do you want know if we are able to anwer them or is this your homework? If you were only sharing interesting problems, and you do love Mathematics, I suggest you to join the group

www.groups.yahoo.com/group/calculusrocks

This is a very interesting forum where you can discuss all the doubts that you have or share interesting problem like these

I will later try to solve them, the first one seems pretty interesting

Take care

Ana

Answer 5

I see your combination up - A = PiR^2 real? nicely, section = Pi (radius squared), the place you placed the radius as 40 8. The Diameter is 24 cm, real? nicely, interior the formulation, radius is only a million/2 of the diameter, so which you need to be utilising 12^2, no longer 40 8^2 Your formulation ought to look like this: section = Pi(3.14) * 12^2 That'll provide you the superb answer.