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Siustrulka
Apr 17, 2008, 07:03 PM
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
Apr 18, 2008, 05:17 AM
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.

.

bdsatish
Apr 23, 2008, 11:32 PM
Good!