Here is a problem that I solved but I don't really know what's the rule??
For the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?
1) f(x)=x^2
2) f(x)=x+1
3) f(x)=square root of x
4) f(x)=2/x
5) f(x)= -3x

I think the best way is to solve by backtracking..

The best way to solve is back-tracking...
f(x)=-3x

f(a)=-3a f(b)=-3b f(a+b)=-3(a+b), f(a)+f(b)=-3(a+b)

so 5 is the answer

similarly we can ssolve other options
f(x) = 2/x
f(a) = 2/a, f(b) = 2/b, f(a+b) = 2/(a+b), f(a)+f(b) = 2*(a+b)/(ab)

f(x) = x+1
f(a) = a+1, f(b) = b+1, f(a+b) = a+b+1, f(a)+f(b) = a+b+2

salma wrote:Here is a problem that I solved but I don't really know what's the rule??
For the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?
1) f(x)=x^2
2) f(x)=x+1
3) f(x)=square root of x
4) f(x)=2/x
5) f(x)= -3x

The answer is 5, what is the best way to solve this problem?

If you do a search, you should find a discussion on this problem.

What we're really being tested on is the concept of distribution. Some operations are distributive, some aren't.

Of the answer choices, the only distributive operation listed is multiplication, which is choice (E).

In other words, for all numbers x, a and b, x(a+b) = xa + xb, which is exactly what the question is looking for.

The same does not hold true for (1) exponents; (2) addition; (3) roots; or (4) division, which is why choices (1)-(4) are wrong.

My second question is as follows:

If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?

10
11
12
13
14

The correct answer is 11.

This is a prime number question. Let's start by breaking down 990:

990 = 10 * 99 = 2 * 5 * 9 * 11

So, for our number to be a multiple of 990, it must include those 4 prime factors. Since the biggest prime on the list is 11, 11! is the smallest factorial that will be a multiple of 990.

("The product of all the integers from 1 to n, inclusive" is the definition of n factorial.)

thanks, stuart.