math problems that i don't get

Test Preparation
1

<p>if x and y are positive integers and 4(2^x)=2^y, what is x in terms of y?
a. y-2
b. y-1
c. y
d. y+1
e. y+2</p>

<hr>

<p>how many integers in the set of all integers from 1 to 100, inclusive, are not the </p>

<h2>square of an integer?</h2>

<p>the acme plumbing company will send a team of 3 plumbers to work on a certain job. </p>

<p>the company has 4 experienced plumber and 4 trainees, if a team consist of 1 experienced </p>

<h2>plumber and 2 trainees, how many different such team are possible?</h2>

<p>if P,R and S are 3 different prime numbers greater than 2, and N=PxRxS, how many </p>

<h2>positive factors including 1 and n does n have?</h2>

<p>can anybody tell me how to do them?</p>

2

<h1>1. You have to know the laws of exponents here and you have to not be intimidated by the algebra. After that, it's simple:</h1>

<p>First, get the 2 exponents on the same side so you can divide and simplify them, like so:</p>

<p>4(2^x)=2^y
divide both sides by (2^x)</p>

<p>4=2^y/(2^x)
law of exponents: when dividing the same base, subtract the denominator's power from the numerator's power</p>

<p>so 2^y/(2^x) = 2^(y-x)</p>

<p>Now express the left side of the equation (4) as a power of 2 (2^2)
and you get
2^2 = 2^(y-x)
now you have a true equation that you can solve for y</p>

<p>y-x = 2 (notice I reversed order just for readability)</p>

<p>-x = 2 - y
multiply by -1
x = y-2
You've now expressed x in terms of y. </p>

<p>the answer is A, y-2.</p>

<h1>2</h1>

<p>The key here is to not be intimidated by 100 integers. Your first reaction might be "gee, how am I ever going to figure all those numbers out?" but if you know that the highest exponent in the closed interval 1 to 100 is 10 (because 10^2 = 100), then you know the only numbers you have to deal with squaring are 10 and less.</p>

<p>In this case, it is all the numbers 1-10 that lead to perfect squares
1 1
2 4
3 9
4 16
5 25
6 36
7 49
8 64
9 81
10 100</p>

<p>The answer is 100-10 = 90. </p>

<p>Notice that I could have immediatly said there was 10, and would have been right, but because 4 is itself a perfect square, I wasn't sure whether to count it once or twice... Turned out fine, but I erred on the same of caution and wrote them all out to be sure.</p>

<p>Your last two questions have been posted on CC before and I'm pretty sure you can find them in the consolidated list of solutions. If not, you can search for them.</p>

<p>Andre</p>

3

<p>thanks a lot ~</p>

4

<p>A quicker way to solve #1 would be</p>

<p>LHS is 4(2^x) = (2^2)(2^x) = 2^(2+x), and RHS = 2^y
So, 2+x = y .</p>

5

<p>Ok there is no need for all those long formulas. The BEST and the QUICKEST way to solve it would be to pick numbers.</p>

<p>Say x=2
y=4</p>

<p>4(2^2)=2^4 WORKS!</p>

<p>so x is y-2.</p>

6

<h1>3 is 4 * 4 * 3 i think.</h1>

<p>the first 4 is for the 4 different experienced plumbers</p>

<p>the second 4 is for the first trainee</p>

<p>the 3 is for the last trainee, since if there are 4 trainees, the second one can't be the same as the first.</p>

<h1>4</h1>

<p>pick numbers for p r and s
p = 3
r = 5
s = 7
3 * 5 * 7 = 105</p>

<p>do factors for 105
1,105
3,35
5,21
7,15</p>

<p>8 factors.</p>

7

<h1>3 is 24. 99% sure.</h1>

8

<p>thecalccobra: Yes, your solution is the quickest. But to a math person, the solution that gives you the fast and easy answer is not always the best. Your solution is probably the best one to use on the SAT because it's quick and foolproof, but it's not the best solution to this problem, as it's kind of a cop-out and only works if you have the multiple-choice answers at your disposal.</p>

9

<h1>1. You have to know the laws of exponents here and you have to not be intimidated by the algebra. After that, it's simple:</h1>

<p>First, get the 2 exponents on the same side so you can divide and simplify them, like so:</p>

<p>4(2^x)=2^y
divide both sides by (2^x)</p>

<p>4=2^y/(2^x)
law of exponents: when dividing the same base, subtract the denominator's power from the numerator's power</p>

<p>so 2^y/(2^x) = 2^(y-x)</p>

<p>Now express the left side of the equation (4) as a power of 2 (2^2)
and you get
2^2 = 2^(y-x)
now you have a true equation that you can solve for y</p>

<p>y-x = 2 (notice I reversed order just for readability)</p>

<p>-x = 2 - y
multiply by -1
x = y-2
You've now expressed x in terms of y.</p>

<p>the answer is A, y-2.</p>

<p>OR</p>

<p>Say x=2
y=4</p>

<p>4(2^2)=2^4 WORKS!</p>

<p>so x is y-2.</p>

<p>you tell me which is quicker and simpler. No need to know any laws of exponents or whatsoever.</p>

10

<p>Sly Si, that question would never appear as a grid-in, only as a multiple choice.</p>

11

<p>The first one is A </p>

<p>The second one deals with combinations ....
( 4 choose 1) * (4 choose 2)
(^You can do this on your calc)
(4 choose 1) = 4
(4 choose 2)= 6
4*6 =24 </p>

<p>And the last one I believe is 8 .....</p>