Question

complicated math for brain exercise !!

The equation a^5 + b^2 + c^2 = 2010 -- a, b, c positive integers -- has a solution in which b and c have a common factor d>1. Find d

ANSWER: d = 7

Answer 1

LaTeX:\[\text{The equation }a^{5}+b^{2}+c^{2}=2010\text{ (}a,b,c\text{ positive integers)}\]\[\text{has a solution in which }b,c,\text{ have a common factor }d\gt1.\text{ Find }d\]

Answer 2

a,b,c are positive

IF a becomes 5 then a^5 will be 3125 which is greater than 2010
so a has to be less than 5

next lets say \(b=dk_1\) and \(c=dk_2\)
so

\(a^5+b^2+c^2=2010\)
\(a^5+d^2(k_1^2+k_2^2)=2010\)

plug in values for a which can rage from 1-4 and then subtract a^5 from 2010
prime factorize the number that you get after this subtraction and in the prime factorization there must be a factor which is a perfect square..that factor will be d

Answer 3

just an approach

Answer 4

hmm okay, that makes sense

Answer 5

Let f(x) equal:\[\frac{\sqrt{x^{2}-1}}{x}\]Find the set of all real values of \(x\) for which \(f(f(x))\) exists.
ANSWER: no values of \(x\)

Why is that?

Answer 6

first off f(x) can be written as \(\sqrt{\frac{x^{2}-1}{x^2}}\) and f(x) is defined for \(x \in (-\infty,-1] \cup [1, \infty)\)
and the range of f(x) is \([0, 1)\)

in f(f(x)) the f(x) inside the bracket part is doing the part of x which must lie in this range \((-\infty,-1] \cup [1, \infty)\) but f(x) can only take values in [0,1) fo f(f(x) ) is not defined

Answer 7

ah, I see, alright thank you :-)



"A rectangle R has width 4 and length 6. The curve C consists of all points outside of R whose distance to the nearest point of R is 1, and D consists of all points outside of C whose distance to the nearest point of C is 1. Find the area enclosed by D, rounded to the nearest square unit" (ANSWER: 77)

Answer 8

What is "the area enclosed by D?" This wording confuses me...