Series of Question 4 - From Guide #4

[sendalot]

P. 44 - Remainder formula
P. 135 - 8 2nd Solutoin??
P. 135 - 10 why 0.5 ??
P. 169 - 8 = ??
P. 169 - 9 = ??
P. 171 - 18
P. 178 - 2 Error for the answer??
P. 178 - 5 Why is it 98 in the solution??

[michael.k.bilow]

Number properties problems can be especially challenging, because nothing we do in our daily lives prepares us for them--only a very select few of us have had to factor a number into primes regularly.

Guide 4, Page 44

It can be a little bit tricky to find the remainder when given two numbers. Here's a trick for doing it:

Suppose you're given the numbers 1973 and 48 and you want to find the remainder that results from dividing them. If you punch 1973/48 into your calculator, you'll get:

1973/48 = 41.10416666

This tells us that the quotient of the two numbers is 41. To find the remainder, multiply the quotient by the smaller of the two numbers, and subtract it from the larger one.

1973 - 48*41 = 5.

Alternatively, 1973 = 48*41 + 5, so the quotient is 5.

Guide 4, Page 135 #8
A, B, and C all lie on a line. D is the midpoint of AB and E is the midpoint of BC. AB = 4 and BC = 10. Which of the following could be the length of AE?

A) 1
B) 2
C) 3
D) 4
E) 5

I really love the explanation of this problem on page 137--read it over very carefully.

The trick to these abstract number line problems is drawing them out correctly. We know that A, B, and C live on a line, and that AB = 4 and BC = 10. We don't, however, know the relationship between A and C, so that gives us two possibilities. The distances are in bold.

1) A--4--B--10--C
or
2) B--4--A--6--C.

Note that both 1) and 2) satisfy the conditions given in the problem, that is, AB = 4 and BC = 10.

There's nothing in the problem so far that forces us to pick either option 1 or option 2, so we have to check both. If we fill in D as the midpoint of AB and E as the midpoint of BC, we have

1) A--2--D--2--B--5--E--5--C
or
2) B--2--D--2--A--1--E--5--C

The distance between BD = AD = 2, and BE = CE = 5 because D and E are the midpoints of AB (= 4) and BC (= 10), respectively. Reading off this diagram, the possible distances for AE are 9 (option 1) and 1 (option 2). Since 1 is the only answer choice that matches, the correct answer is A.

Guide 4, Page 135 #10
S is the midpoint of QR on the number line. R = -2Q
QC:
Quantity A: S
Quantity B: 0

Plenty of ways to do this problem. Picking numbers works well, but pure algebra isn't that hard for this one, either.

Let's convert "S is the midpoint of QR" into an algebra statement. This one should definitely be in your toolkit for the test. Ready?

S = (Q+R)/2.

Now, we have that R = -2Q, so

S = (Q -2Q)/2 = -Q/2.

The hardest part of this problem is proving that Q must be negative. Remember that one of the things you can assume on the GRE is that the points on a line are in the correct order. Thus if Q is to the left of R on the number line, we can safely assume that Q < R.

Q = -2R implies that Q and R have opposite signs. Because R > Q, Q must be negative. Q cannot be zero, because the inequality is strict. Therefore, S = -Q/2 is the opposite of a negative number, and must be positive. The correct answer is therefore A.

Guide 4, Page 169 #8
Between 5 and 10 on a number line, exclusive, a dark gray tick mark is placed at every multiple of 1/3 and a light gray tick mark is placed at every multiple of 1/9. At how many places will a dark gray tick mark and a light gray tick mark overlap? Fill in the answer.

The key to this problem is realizing that every multiple of 1/3 is also a multiple of 1/9. Therefore, every multiple of 1/3 strictly between 5 and 10 will have both the dark gray and light gray tick mark.

A little trick here is that we are looking at the numbers between 5 and 10 (which are multiples of 1/3), exclusive. That means we need to avoid 5 and 10 when we do our count of multiples of 1/3.

One of the formulas that you must have memorized for the exam is the formula for counting the number of numbers in an inclusive or exclusive set. The formula for the number of integers between two integers x > y, inclusive, is x - y + 1. For example, the number of integers between 2 and 8, inclusive is 7, which is 8 - 2 + 1.

The number of integers between x > y, exclusive is x - y - 1.
We know that 5 = 15/3 and 10 = 30/3, and that all of the integer multiples of 1/3 will be in our set, so the number of these tick marks will be 30 - 15 - 1 = 14, since this set excludes its endpoints. The answer is 14.

Guide 4, Page 169 #9
Suppose that |b-a| = 2. Which of the following statements must be true?

A) a must be positive if b is positive.
B) a must be negative if b is negative.
C) b > 0 if a > 2.


I think testing numbers is the best way to solve this problem, but we have to be smart about the numbers we choose to test. Remember, the ranges you should pull from are:

x>1
0 < x < 1
-1 < x < 0
x < -1
x = {0,1,-1}

I won't go through testing each range, but a trick to notice here is that because the question is asking about positives and negatives, you'll do better by testing values close to zero.

A is false.
Let b = 1. Then |b-a| = 2 means a = -1 or a = +3. Since (a,b) = (1,-1) satisfies the equation, it is not always true that if b is positive, that a must also b positive.

Similarly, B is false.
Let b = -1. It follows that a = +3 or -1, which means that if b is negative, a doesn't need to be negative as well.

By process of elimination, we know that C must be true, since each GRE question will always have at least one correct answer.

No matter what the value of a is, the possible values for b will be a+/-2. We can test this out--|b-a| = 2 implies b-a = 2 or b-a = -2, from which we get b = a +/- 2. We want to know what b *must* be larger than, so we can focus on the small value, b = a-2.

Now let's plug in the condition a > 2. Then b > 2 - 2, which is equivalent to saying b > 0, which is exactly condition C.


Guide 4, Page 171 #18
Points X, Y, Z, and W lie on a number line, but not necessarily in that order. The distance between X and Y is 10, the distance between Z and W is 7, and the distance between Y and Z is 4. What is the minimum distance between X and W.

Here's a quick way to do the problem: The distance between X and Y is 10, and the distance between Y and W is either 11 or 3. Since there is no restriction on the order, we can put W 11 units away from Y on the same side as X, which is 10 units away. Therefore, the minimum distance between X and W is 1.

Let's do it a little bit more deliberately. It will be easiest for us to build this answer up piece by piece.

Let's start by drawing XY. The distance is in bold.

X--10--Y

There's no other way to draw this, so we can breathe easy.

Now let's add in the distance between Y and Z. We have two options. Z could be between Y and X, or not.

1) X--6--Z--4--Y
2) X--10--Y--4--Z

Option 1 looks like it has a better chance of putting W near X, but we can't be positive yet. Let's explore all the possibilities, given that ZW = 7.

1a) W--1--X--6--Z--4--Y
1b) X--6--Z--4--Y--3--W
2a) X--7--W--3--Y--4--Z
2b) X--10--Y--4--Z--7--W

Checking all four possible options, it's clear that 1a minimizes the size of WX at 1. The correct answer is A.

Book 4, Page 178 #2
QC:
Quantity A: (6^8*5^2*4^2)/(3^6*8^2)
Quantity B: (6^9*5^3*4^3)/(3^8*2^8)


There is a typo in the answer given on page 182. The correct answer is B.
These two numbers have a lot in common, and we can do anything to a QC that we can legally do to an inequality, for example, multiplying or dividing by a positive number.

Let's start by dividing both A and B by the entire numerator of Quantity A. This is a positive number, so we don't need to worry about making a mess of the inequality. Also, if we divide the numerator of Quantity A by itself, we'll just be left with 1, which makes the problem a lot simpler. Looking at the numerator of Quantity B, we can see that it is the exact same as quantity A, except that the exponents are 1 bigger. Dividing the numerator will just leave 6*5*4.

After dividing both quantities by the numerator of quantity A (6^8*5^2*4^2):

Quantity A: 1/(3^6*8^2)
Quantity B: (6^1*5^1*4^1)/(3^8*2^8)

Already, we're much simpler. Let's get rid of that 8 in the bottom of quantity A. 8 = 2^3, so 8^2 = (2^3)^2 = 2^(3*2) = 2^6.

Now, we have:

Quantity A: 1/(3^6*2^6)
Quantity B: (6^1*5^1*4^1)/(3^8*2^8)

To simplify this even further, let's multiply both sides by the denominator of Quantity B. Just as before, we're multiplying by a positive number, so there's nothing in the inequality to worry about. The denominator of quantity B will disappear, and just by looking, we can see that the

So, after multiplying by the denominator of quantity B (3^8*2^8), we have:

Quantity A: 3^2*2^2
Quantity B: (6^1*5^1*4^1)

Now we can just multiply these numbers out.

Quantity A: 36
Quantity B: 120

Therefore Quantity B is always larger.[\b]

[b]Book 4, Page 178 #5
M and N are positive 2-digit integers and N = (M+5)/2. If M/3 is not an integer, what is the largest possible value of N?

We know that M must be odd (otherwise (M+5)/2 would not be an integer). Since N is about half of M, the larger M is, the larger N will be. So let's find the largest possible acceptable value of M.

The largest 2-digit integer is 99, but 99/3 = 33 is an integer, so we can't use it.

The next largest two digit integer, 98, is even, so that won't work either.

Going down the line, 97/3 = 32.333... is not an integer, and 97 is odd. Bingo. N = (97 + 5)/2 = 102/2 = 51.

Great questions!

Michael