Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**sureshrgp****Member**- Registered: 2006-06-07
- Posts: 11

1) Given that f(x) is a polynomial of degree EIGHT such that

f(m) = 1/m for m = 1,2,3,4,5,...,9

Find f(10)

2) Suppose a²+ab+b²-c²-cd-d²=0, where a, b, c, d are positive integers.

Then I think a+b+c+d is a composite number. It is true? Prove it!

3) Prove that if p and p²+8 are prime then so is p³+4.

4) In how many ways can you change a $50 bill using $20 bills, $10 bills,

$5 bills, $2 bills, and $1 bills?

5) Find the equation of the circle which touches the axis of y at a distance 4

from the origin and cuts off an intercept of length 6 on the axis of x.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

1) Given that f(x) is a polynomial of degree EIGHT such that

f(m) = 1/m for m = 1,2,3,4,5,...,9

Find f(10)

This is a bit like taylor series. Since it's of degree 8, you know it looks like:

y = ax^8 + bx^7 + cx^6 + ... + gx^2 + hx + i

So start off with 1:

1/1 = a + b + c + d + e + f + g+ h + i, since all x's are 1.

Continue this for 2, 3, 4, 5, and so on. Then just solve the system of equations...

That seems to be too difficult. There is probably an easier way, but I'm not sure what it is.

2)

You know that (a + b)^2 = (c + d)^2. Thus, |a + b| = |c + d|. But a, b, c, and d are all positive, so a + b = c + d. Thus, you know that a + b + c + d = 2(a + b), and so 2 divides a + b + c + d, thus, it's composite.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**sureshrgp****Member**- Registered: 2006-06-07
- Posts: 11

Thanks!!

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

5.) (x - 5)² + (y - 4)² = 25

**igloo** **myrtilles** **fourmis**

Offline