Siustrulka Posts: 33, Reputation: 1 Junior Member #1 Apr 17, 2008, 07:03 PM
integral factors
what is the total number of positive integral factors of (60)^5

A point (x,y) is called integral if both x and y are integers. How many points on the graph of 1/x + 1/y=1/4 are integral points?

The complex numbers 1+i and 1+2i are both roots of the equation
x^5- 6x^4+Ax^3+Bx^2+Cx+D=0
where A,B,C, and D are integers. What is the value of A+B+C+D?
 galactus Posts: 2,271, Reputation: 282 Ultra Member #2 Apr 18, 2008, 05:17 AM
Originally Posted by Siustrulka
what is the total number of positive integral factors of (60)^5

$60^{5}=777,600,000=2^{10}\cdot{3^{5}}\cdot{5^{5}}$

If you mean how many divisors does 60^5 have, then you can make an observation from basic number theory. When you factor a number you get the form $2^{a}3^{b}5^{c}..........$. The number of divisors is given by looking at the exponents of the prime factors.
The number of divisors is (a+1)(b+1)(c+1)...
So, 60^5 has (11)(6)(6) divisors. A divisor is a number which divides evenly into a number N.

A point (x,y) is called integral if both x and y are integers. How many points on the graph of 1/x + 1/y=1/4 are integral points?
How many ways can you find two fractions which sum to 1/4?

You could solve the equation for y and then plug in integer values of x and see how many return integer values of y.

Here are a few: (2,-4), (3,-12), (5,20),.

The complex numbers 1+i and 1+2i are both roots of the equation
x^5- 6x^4+Ax^3+Bx^2+Cx+D=0
where A,B,C, and D are integers. What is the value of A+B+C+D?
This quintic equation has 5 roots. Since 1+i and 1+2i are roots, their conjugates are also roots. So, 4 roots are 1+i, 1+2i, 1-i, 1-2i

We have one other root we must find since there must be 5 rrots and we have 4.

Using our given roots we have a quartic: $(x-(1+i))(x-(1-i))(x-(1+2i))(x-1-2i))=x^{4}-4x^{3}+11x^{2}-14x+10$

This quartic multiplied by the other root, x-y, will give us our given quintic:

$(x-y)(x^{4}-4x^{3}+11x^{2}-14x+10)=x^{5}-6x^{4}+Ax^{3}+Bx^{2}+Cx+D$

$x^{5}-yx^{4}-4x^{4}+4yx^{3}+11x^{3}-11yx^{2}-14x^{2}+14yx+10x-10y=x^{5}-6x^{4}+Ax^{3}+Bx^{2}+Cx+D$

Equate coefficients:

$-y-4=-6$

$4y+11=A$

$-11y-14=B$

$14y+10=C$

$-10y=D$

Now, solve the easy system by solving the first one for y and subbing in the others, then add up your solutions.

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 bdsatish Posts: 2, Reputation: 1 New Member #3 Apr 23, 2008, 11:32 PM
Comment on galactus's post
Good!

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